<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

546. Ternary Password
Time limit per test: 1 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



In the ternary world all passwords are ternary, that is, they consist only of digits "0", "1" and "2". Terentius is trying to register on a famous internet service site Toogle, but the problem is, according to the security policy of this service the password must contain exactly a characters "0" and exactly b characters "1". All other characters of the password must be "2".

Terentius was fond of his password, he spent much time trying to remember it and now he can type it even with his eyes closed. That's the reason Terentius wants to replace the minimum number of characters in his password so that it meets the strict requirements of the Toogle policy. Terentius wants only to replace (substitute) some characters in password, he doesn't intend to perform other operations with the password.

Help Terentius find the minimum number of replacements and print the corresponding possible variant of the password to Toogle.

Input
The first line of the input contains three integers n, a and b (1 ≤ n ≤ 200; 0 ≤ a,b ≤ 200) — the length of Terentius's password, the expected number of characters "0" and the expected number of characters "1" in the password to Toogle. The second line contains Terentius's password. All characters in this line are digits "0", "1" or "2".

Output
In the first output line print t — the minimum number of replacements. In the second line print the possible variant of the Toogle password — the password that satisfies the Toogle safety policy, that differs from Terentius's password in exactly t positions. If there are several passwords, print any of them. Obviously, the length of the printed password must equal n.

It is possible that Terentius's password already meets the Toogle policy. In this case the first line must contain "0", and the second line must contain Terentius's password.

If the solution doesn't exist, that is, if it is impossible to get a password to Toogle if Terentius uses replacements only, then print "-1" in the first line of the output. In this case print empty second line or don't print the second line at all.

Example(s)
sample input
sample output
6 1 3
012022
2
111022

sample input
sample output
5 5 0
02211
4
00000

<|response|>
1. Abridged problem statement
Given a ternary string s of length n (characters '0', '1', '2') and targets a (number of '0') and b (number of '1'). You may only substitute characters; you cannot insert or delete. Find the minimum number of substitutions to make the string contain exactly a zeros, exactly b ones, and hence n – a – b twos. Output that minimum and one resulting string, or –1 if a + b > n.

2. Key observations
- Let c0, c1, c2 be the counts of '0', '1', '2' in the original string.
- Define deficits/surpluses
  d0 = a – c0 (positive means need more zeros, negative means too many zeros)
  d1 = b – c1 (analogous for ones)
- You can "swap" a surplus of one digit directly into a deficit of another:
  Convert some '1'→'0' if d0>0 and d1<0, or '0'→'1' if d1>0 and d0<0.
  Each such direct swap fixes two imbalances at the cost of one replacement.
- After maximal direct swaps, at least one of d0 or d1 becomes zero or both have the same sign.
- Remaining imbalances must be resolved by converting to/from '2':
  If d0>0, convert d0 copies of '2'→'0'. If d0<0, convert –d0 copies of '0'→'2'.
  Similarly for d1 and '1'.
  Each such conversion costs |d0|+|d1| replacements.

3. Full solution approach
Step A: Feasibility check
  If a + b > n, print "-1" and exit.
Step B: Count existing digits and compute deficits
  Scan s once to compute c0, c1, c2, then set d0 = a – c0, d1 = b – c1.
Step C: Direct swaps between '0' and '1'
  1. If d0>0 and d1<0, let q = min(d0, –d1). Replace q occurrences of '1'→'0', update d0–=q, d1+=q, cost += q.
  2. If d1>0 and d0<0, let q = min(d1, –d0). Replace q occurrences of '0'→'1', update d1–=q, d0+=q, cost += q.
Step D: Fix remaining imbalances via '2'
  - Total extra cost = |d0| + |d1|.
  - Traverse s again, for each character:
    If d0<0 and char=='0': change to '2', d0++.
    Else if d0>0 and char=='2': change to '0', d0–.
    Else if d1<0 and char=='1': change to '2', d1++.
    Else if d1>0 and char=='2': change to '1', d1–.
  Continue until d0==0 and d1==0.
Step E: Output total cost and the modified string.

4. C++ implementation with comments

```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;
};

int n, a, b;
string s;

void read() {
    cin >> n >> a >> b >> s;
}

void solve() {
    // We need exactly a zeros and b ones; everything else becomes a two.
    //
    // Subtract the zeros and ones already present from a and b, so a and b
    // become the signed deficits: a positive value means we still need that
    // many of the digit, a negative value means we have that many in excess.
    //
    // If a + b > n there is no way to fit the required zeros and ones, so the
    // answer is -1.
    //
    // A surplus zero (a < 0) can directly cover a needed one (b > 0) by a
    // single relabel, and vice versa; each such swap fixes two deficits with
    // one replacement, so we greedily apply those first. Whatever deficit
    // remains is settled against the twos: each leftover replacement turns a
    // surplus 0/1 into a 2 or a 2 into a needed 0/1. The total number of
    // replacements is the sum of the remaining absolute deficits plus the
    // swaps already made.

    if(a + b > n) {
        cout << -1 << '\n';
        return;
    }

    for(char c: s) {
        if(c == '0') {
            a--;
        } else if(c == '1') {
            b--;
        }
    }

    int answer = 0;
    if(a > 0 && b < 0) {
        int q = min(a, -b);
        answer += q;
        b += q;
        a -= q;
        for(char& c: s) {
            if(c == '1' && q) {
                c = '0';
                q--;
            }
        }
    }

    if(b > 0 && a < 0) {
        int q = min(-a, b);
        answer += q;
        b -= q;
        a += q;
        for(char& c: s) {
            if(c == '0' && q) {
                c = '1';
                q--;
            }
        }
    }

    answer += abs(a) + abs(b);

    for(char& c: s) {
        if(a < 0 && c == '0') {
            a++;
            c = '2';
        } else if(a > 0 && c == '2') {
            a--;
            c = '0';
        } else if(b < 0 && c == '1') {
            b++;
            c = '2';
        } else if(b > 0 && c == '2') {
            b--;
            c = '1';
        }
    }

    cout << answer << '\n';
    cout << s << '\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;
}
```

5. Python implementation with comments

```python
import sys

def main():
    data = sys.stdin.read().split()
    n, a, b = map(int, data[:3])
    s = list(data[3])

    # Impossible if we need more 0s+1s than length
    if a + b > n:
        print(-1)
        return

    # Subtract existing counts
    for c in s:
        if c == '0':
            a -= 1
        elif c == '1':
            b -= 1
    # Now a>0 means we still need zeros; a<0 means we have surplus zeros.
    # Similarly for b and ones.

    ans = 0

    # 1) Directly swap surplus '1' into needed '0'
    if a > 0 and b < 0:
        # We can perform up to min(a, -b) swaps
        q = min(a, -b)
        ans += q
        a -= q
        b += q
        # Apply swaps in the string
        for i, c in enumerate(s):
            if q == 0:
                break
            if c == '1':
                s[i] = '0'
                q -= 1

    # 2) Directly swap surplus '0' into needed '1'
    if b > 0 and a < 0:
        q = min(b, -a)
        ans += q
        b -= q
        a += q
        for i, c in enumerate(s):
            if q == 0:
                break
            if c == '0':
                s[i] = '1'
                q -= 1

    # 3) Remaining deficits or surpluses must be handled via '2'
    # Each unit of imbalance costs 1 replacement
    ans += abs(a) + abs(b)

    # Apply conversions with '2'
    for i, c in enumerate(s):
        if a < 0 and s[i] == '0':
            # surplus zero → convert to '2'
            s[i] = '2'
            a += 1
        elif a > 0 and s[i] == '2':
            # need more zero → convert '2' to '0'
            s[i] = '0'
            a -= 1
        elif b < 0 and s[i] == '1':
            # surplus one → convert to '2'
            s[i] = '2'
            b += 1
        elif b > 0 and s[i] == '2':
            # need more one → convert '2' to '1'
            s[i] = '1'
            b -= 1

    # Print result
    print(ans)
    print(''.join(s))


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