## 1) Abridged problem statement

For each test case, given an odd prime \(P\) (\(1<P<1000\)), consider the set \(\{1,2,\dots,2P\}\). Count how many ways to choose **exactly \(P\) distinct numbers** such that their sum is divisible by \(P\). Output the count for each test.

---

## 2) Detailed editorial (solution explanation)

### Key observation
We are counting \(P\)-element subsets \(S \subset \{1,\dots,2P\}\) such that
\[
\sum_{x\in S} x \equiv 0 \pmod P.
\]
The total number of \(P\)-subsets is \(\binom{2P}{P}\), which is huge, so we need a number-theoretic counting argument rather than DP.

### Two “special” valid subsets
Split the numbers into two blocks:
- Small block: \(A=\{1,2,\dots,P\}\)
- Large block: \(B=\{P+1,P+2,\dots,2P\}\)

Both are size \(P\). Their sums are:
- \(\sum_{i=1}^P i = \frac{P(P+1)}{2}\), divisible by \(P\).
- \(\sum_{i=P+1}^{2P} i = \sum_{i=1}^{2P} i - \sum_{i=1}^P i\) is also divisible by \(P\).

So at least **2** subsets are good: \(A\) and \(B\).

### Grouping the remaining subsets into blocks of size \(P\)
Now take any other \(P\)-subset \(S\) (so \(S\neq A\) and \(S\neq B\)). Let:
- \(k = |S \cap A|\), the count of chosen numbers from \(\{1,\dots,P\}\).

Because \(S\) is not all-small and not all-large, we have:
\[
1 \le k \le P-1.
\]

#### Define a “shift” operation on the small elements
For each \(i \in \{0,1,\dots,P-1\}\), create a new subset \(S_i\) from \(S\) by:
- Keeping all chosen numbers in \(B\) unchanged.
- Replacing each chosen number \(x \in S\cap A\) with
  \[
  x' = ((x-1+i)\bmod P)+1,
  \]
  i.e., shift within \(\{1,\dots,P\}\) cyclically.

This produces \(P\) subsets \(S_0,S_1,\dots,S_{P-1}\), all of size \(P\).

#### These \(P\) subsets are all distinct
Shifting by different \(i\) yields different sets of small elements (cyclic shift on residues mod \(P\)), and the large elements are fixed, so \(S_i\) differ.

So we have a group (“orbit”) of exactly \(P\) distinct subsets.

### How the sum modulo \(P\) changes
When we shift the \(k\) small elements by \(i\), each small element increases by \(i\) modulo \(P\), so the total sum increases by:
\[
k\cdot i \pmod P.
\]
Therefore,
\[
\sum(S_i) \equiv \sum(S_0) + k i \pmod P.
\]

### Why exactly one subset per group is good
Since \(P\) is prime and \(1\le k \le P-1\), we have \(\gcd(k,P)=1\). Thus multiplication by \(k\) permutes residues modulo \(P\). As \(i\) runs from \(0\) to \(P-1\), the values \(k i \bmod P\) hit every residue exactly once. Hence the \(P\) sums \(\sum(S_i) \bmod P\) are all different and cover all residues, so **exactly one** of them is congruent to \(0\).

So: among these \(P\) subsets, exactly **1** is good.

### Counting
- Total subsets: \(\binom{2P}{P}\)
- Special ones: 2 good subsets (\(A\), \(B\))
- Remaining subsets: \(\binom{2P}{P}-2\), partitioned into groups of size \(P\)
- Each group contributes exactly 1 good subset

So the total number of good subsets is:
\[
2 + \frac{\binom{2P}{P}-2}{P}.
\]

### Big integer need
For \(P\) up to 999, \(\binom{2P}{P}\) is enormous, so we must use big integers. The provided C++ code includes a full big integer implementation and computes:
1. \(C=\binom{2P}{P}\) via multiplicative formula
2. Answer \(= 2 + \frac{C-2}{P}\)

---

## 3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int p;

void read() { cin >> p; }

void solve() {
    // There is a classic DP solution but that would be too slow, especially as
    // we need big integers. It's possible to precompute all answers which might
    // be doable as there are 1000 unique inputs, but that solution isn't
    // interesting.
    //
    // Instead, because P is an odd prime we have nice properties. Let's first
    // consider the two sets {1, ..., P} and {P+1, ..., 2P} which already have
    // sums divisible by P. Now we can consider all other subsets (there are
    // C(2*P,P) - 2 of them). We can make groups of exactly P subsets each:
    //
    //   Take any such subset. Let k = how many of its numbers
    //   are from 1 to P (so 1 <= k <= P-1).
    //
    //   Now create P versions by adding the same number i (0 to P-1)
    //   to each of those k small numbers, then wrap around mod P
    //   so they stay between 1 and P. The big numbers stay the same.
    //
    //   This gives P different P-sized subsets. Their sums change
    //   by k*i each time. Since k and P are coprime, k*i mod P
    //   hits every possible remainder exactly once as i goes 0..P-1.
    //
    //   Therefore exactly one subset in each team of P has sum
    //   divisible by P.
    //
    // So the ordinary subsets contribute (C(2*P,P) - 2) / P good ones.
    // Adding the two special ones gives 2 + (C(2*P,P) - 2) / P.
    //
    // This is explanation is heavily based on the below and it turns out this
    // problem also appeared at IMO 1995:
    //
    //     https://math.stackexchange.com/questions/314788/let-p-be-an-odd-prime-
    //             number-how-many-p-element-subsets-of-1-2-3-4-ldo

    bigint res = 1;
    for(int i = 0; i < p; ++i) {
        res *= (2 * p - i);
        res /= (i + 1);
    }
    res -= 2;
    res /= p;
    res += 2;
    cout << res << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same logic) with detailed comments

```python
import sys
import math

def solve():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    t = int(data[0])
    out = []
    idx = 1

    for _ in range(t):
        p = int(data[idx]); idx += 1

        # Compute C(2p, p) as an exact integer.
        # Python ints are arbitrary precision, so no custom bigint needed.
        #
        # Multiplicative formula:
        # C(2p, p) = Π_{i=1..p} (p+i)/i = Π_{i=0..p-1} (2p-i)/(i+1)
        c = 1
        for i in range(p):
            c *= (2 * p - i)
            c //= (i + 1)   # exact division at every step

        # Apply derived counting formula:
        # answer = 2 + (C(2p,p) - 2) / p
        ans = 2 + (c - 2) // p
        out.append(str(ans))

    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

Let \(P\) be an odd prime. Two \(P\)-subsets \(\{1,\dots,P\}\) and \(\{P+1,\dots,2P\}\) have sums divisible by \(P\). Any other \(P\)-subset has \(k\in[1,P-1]\) elements from \(\{1,\dots,P\}\). Cyclically shifting those \(k\) small elements by \(i\in\{0,\dots,P-1\}\) forms an orbit of exactly \(P\) distinct subsets; their sums change by \(k i \pmod P\). Since \(\gcd(k,P)=1\), \(k i\) runs through all residues mod \(P\), so exactly one subset in each orbit has sum \(\equiv 0\pmod P\). Thus good subsets:
\[
2 + \frac{\binom{2P}{P}-2}{P}.
\]
Compute \(\binom{2P}{P}\) with big integers and output the formula.