## 1) Concise abridged problem statement

There are `n` groups; group `i` has `a[i]` friends. The cafe has identical tables with `r` seats each. You must seat everyone using the minimum number of tables, with the constraint:

- No person may sit alone without a teammate at their table.  
  (Equivalently, every group’s people must be split across tables into parts of size **at least 2**.)

Different groups may share a table. A group may occupy multiple tables. Empty seats are allowed.

Output the minimum number of tables needed.

---

## 2) Detailed editorial (explaining the solution)

### Key observation: only 2’s and 3’s matter
If a group puts `k ≥ 2` people on some table, that chunk size is allowed. Any `k ≥ 2` can be decomposed into chunks of size **2 and 3** (because):
- `2 = 2`
- `3 = 3`
- `4 = 2+2`
- `5 = 2+3`
- `6 = 3+3` or `2+2+2`
- and so on.

So we can think of each group as being split into some number of **pairs (2)** and **triples (3)**.

Let:
- `T` = number of tables we try to use
- `r` = seats per table
- `S` = total people `sum(a[i])`
- `E = T*r - S` = total empty seats (must be `>= 0`)

Now the seating problem becomes: choose how many “3-chunks” exist in total, and see if they can be packed into `T` tables of size `r` along with the remaining “2-chunks”, allowing `E` empty seats.

---

### How many triples (3-chunks) are possible?

For each group size `x`:

- If `x` is even: it can be all pairs, but you can also replace each block of 6 people (`2+2+2`) by `3+3`, increasing the number of triples by **2** each time.  
  So triples can be `0, 2, 4, ..., 2*(x/6)`.

- If `x` is odd: you must have at least one triple (since sum of 2’s is even). After removing one triple, the remaining `x-3` is even and again you can convert each 6 block into `+2` triples.  
  So triples can be `1, 3, 5, ..., 1 + 2*((x-3)/6)`.

Therefore:
- `min_threes` = number of odd groups (each requires at least one triple)
- `max_threes` = sum over groups of the maximum achievable triples by these rules

And any feasible total number of triples must have the same parity as `min_threes` and lie in `[min_threes, max_threes]` stepping by `2`.

---

### Why binary search on the number of tables?
We need the **minimum** number of tables. If seating is possible with `T` tables, it will also be possible with `T+1` tables (more capacity, more empty seats). This monotonicity allows binary search on `T`.

Lower bound: `ceil(S / r)` (you need enough seats)  
Upper bound: `S` (always possible with one person per seat? Actually constraint forbids singles, but with `r>=3` and `a[i]>=2`, `S` is still a safe crude upper bound used in code.)

---

### Feasibility check for a fixed `T`

We iterate over possible total triples `threes` in `[min_threes, max_threes]` step 2.

Given a choice of `threes`:
- People used in triples: `3*threes`
- Remaining people must be seated in pairs: `S - 3*threes` must be `>= 0` and even.

Now we must see if we can pack these triples/pairs into `T` tables of size `r`, possibly leaving empty seats.

This depends heavily on `r` parity:

---

#### Case A: `r` is even
Then each table has even capacity, so pairs fit naturally. Triples create “oddness” that must be handled:

- In 6 seats, you can place either `3+3` (two triples) or `2+2+2` (all pairs).  
  This means each table can “absorb” triples in chunks corresponding to `r/6` blocks of 6 seats:
  - `absorbed = 2 * (r/6) * T` triples can be placed without needing empty seats, by using 6-seat blocks as `3+3`.

Any remaining triples after that cannot be perfectly paired into 6-blocks; to place an unpaired triple on an even-sized table, you effectively waste 1 seat somewhere (empty seat acts as parity-fixer).  
So remaining triples must be `<= E`.

That’s what the code enforces.

---

#### Case B: `r` is odd
Odd table sizes introduce an additional parity issue: filling an odd-sized table using only pairs (even contributions) is impossible unless you leave at least one seat empty or place a triple.

The code handles this by:

1) If we have fewer triples than tables (`threes < T`):  
   some tables would contain no triple. Each such table needs **at least 1 empty seat** to make the used seats even.  
   That requires `E >= (T - threes)`. If yes, feasible immediately.

2) Otherwise (`threes >= T`):  
   put **one triple on every table**, consuming `T` triples and `3` seats per table.  
   Now each table has remaining capacity `r-3` which is even, reducing to the even-`r` situation on leftover triples.

   After this reduction:
   - If `E == 0` and leftover triples count is odd, impossible (cannot resolve parity without empty seats).
   - Use 6-seat absorption as before with `cur_r = r-3`.
   - Any remaining triples must be covered by empty seats `E`.

If any `threes` choice works, `check(T)` returns true.

---

### Complexity
- Computing `min_threes`, `max_threes`: `O(n)`
- Binary search over `T`: `O(log S)` (S ≤ 4e6 in worst case but here ≤ 2000*2000=4e6)
- Each check loops over possible `threes`, which can be up to about `O(S/3)` in worst case but bounded by derived max; in practice it’s fast enough under constraints, and intended solution passes within 1s in C++.

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Helper to print a pair (not used in final output but convenient for debugging).
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

// Helper to read an entire vector: assumes size is already set.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x: a) {
        in >> x;
    }
    return in;
};

// Helper to print a vector (unused here).
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, r;          // n = number of groups, r = seats per table
vector<int> a;     // group sizes

// Read input.
void read() {
    cin >> n >> r;
    a.resize(n);
    cin >> a;
}

/*
Check if it is possible to seat everyone using exactly 'tables' tables.

Parameters:
- tables: number of tables T
- total_people: S = sum(a[i])
- min_threes: minimum number of 3-chunks required (one per odd group)
- max_threes: maximum number of 3-chunks possible overall

The function tries feasible totals of 3-chunks (triples) and decides if a packing exists.
*/
bool check(int64_t tables, int64_t total_people, int64_t min_threes, int64_t max_threes) {
    // Total empty seats if we use 'tables' tables.
    int64_t empty = tables * r - total_people;
    if (empty < 0) {
        // Not enough seats at all.
        return false;
    }

    // Try every possible number of triples, same parity as min_threes, within range.
    for (int64_t threes = min_threes; threes <= max_threes; threes += 2) {
        // After choosing 'threes' triples, remaining people must be in pairs.
        int64_t remaining_seats = total_people - 3 * threes;

        // Cannot use more triples than people, and leftover must be even (pairs).
        if (remaining_seats < 0 || remaining_seats % 2 != 0) {
            continue;
        }

        // Work with mutable copies for further casework.
        int64_t cur_threes = threes;
        int64_t cur_r = r;

        // If table size is odd, we must manage parity.
        if (cur_r % 2 == 1) {
            // If we have fewer triples than tables, some tables would have no triple.
            // Each such table needs at least one empty seat to make occupancy even.
            if (cur_threes < tables) {
                if (empty >= tables - cur_threes) {
                    // We can spend 1 empty seat for each table without a triple.
                    return true;
                }
                // Not enough empty seats to "fix" those tables => try next triple count.
                continue;
            }
            // Otherwise, place one triple on every table:
            // reduce triple count by tables, and reduce table capacity by 3.
            cur_threes -= tables;
            cur_r -= 3; // now cur_r is even
        }

        // Now cur_r is even.
        // If we have zero empty seats and an odd number of triples left,
        // we cannot fix parity in even-sized tables.
        if (empty == 0 && cur_threes % 2 == 1) {
            continue;
        }

        // Each block of 6 seats can absorb 2 triples (3+3).
        // Number of 6-blocks per table is cur_r/6.
        int64_t absorbed = 2 * (cur_r / 6) * tables;

        // If all remaining triples can be placed inside 6-blocks, feasible.
        if (cur_threes <= absorbed) {
            return true;
        }

        // Otherwise, triples that couldn't be absorbed must be handled separately.
        cur_threes -= absorbed;

        // Each leftover triple needs one empty seat somewhere to resolve mismatch.
        if (empty >= cur_threes) {
            return true;
        }
    }

    // No triple count worked.
    return false;
}

void solve() {
    /*
    We precompute:
    - total_people S
    - min_threes: at least 1 triple for each odd group
    - max_threes: maximum triples possible after converting 6-chunks into +2 triples
    */
    int64_t total_people = 0;
    int64_t min_threes = 0;
    int64_t max_threes = 0;

    for (int sz: a) {
        total_people += sz;

        if (sz % 2 == 0) {
            // even sz: triples can be 0,2,4,...,2*(sz/6)
            max_threes += 2 * (sz / 6);
        } else {
            // odd sz: must have at least one triple
            min_threes++;
            // maximum is 1 + 2*((sz-3)/6)
            max_threes += 1 + 2 * ((sz - 3) / 6);
        }
    }

    // Binary search the minimum number of tables.
    int64_t low = (total_people + r - 1) / r; // at least enough seats
    int64_t high = total_people;              // safe upper bound
    int64_t ans = high;

    while (low <= high) {
        int64_t mid = (low + high) / 2;
        if (check(mid, total_people, min_threes, max_threes)) {
            ans = mid;        // mid tables is feasible
            high = mid - 1;   // try fewer
        } else {
            low = mid + 1;    // need more
        }
    }

    cout << ans << '\n';
}

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

    int T = 1;
    // cin >> T; // problem has a single test
    for (int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

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

```python
import sys

def can_seat(T, n, r, total_people, min_threes, max_threes):
    """
    Return True if we can seat everyone using exactly T tables.
    We try all feasible counts of 3-sized subgroups (triples).
    """
    empty = T * r - total_people
    if empty < 0:
        return False

    # Try triple counts with correct parity: min_threes, min_threes+2, ...
    threes = min_threes
    while threes <= max_threes:
        remaining = total_people - 3 * threes

        # Remaining people must be representable by pairs => even and non-negative
        if remaining >= 0 and remaining % 2 == 0:
            cur_threes = threes
            cur_r = r

            # If r is odd, handle parity constraints per table
            if cur_r % 2 == 1:
                if cur_threes < T:
                    # Tables without a triple need 1 empty seat each
                    if empty >= (T - cur_threes):
                        return True
                    threes += 2
                    continue
                # Put one triple on every table so leftover capacity per table becomes even
                cur_threes -= T
                cur_r -= 3  # now even

            # Now cur_r is even
            if empty == 0 and (cur_threes % 2 == 1):
                threes += 2
                continue

            # Each 6 seats can contain two triples (3+3), absorbing 2 triples per 6-block
            absorbed = 2 * (cur_r // 6) * T
            if cur_threes <= absorbed:
                return True

            cur_threes -= absorbed

            # Each remaining triple requires one empty seat to "fix" parity/waste a seat
            if empty >= cur_threes:
                return True

        threes += 2

    return False


def solve():
    data = list(map(int, sys.stdin.read().strip().split()))
    n, r = data[0], data[1]
    a = data[2:2+n]

    total_people = sum(a)

    # Compute min/max possible number of triples overall
    min_threes = 0
    max_threes = 0
    for sz in a:
        if sz % 2 == 0:
            max_threes += 2 * (sz // 6)
        else:
            min_threes += 1
            max_threes += 1 + 2 * ((sz - 3) // 6)

    # Binary search for minimal feasible number of tables
    low = (total_people + r - 1) // r
    high = total_people
    ans = high

    while low <= high:
        mid = (low + high) // 2
        if can_seat(mid, n, r, total_people, min_threes, max_threes):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    print(ans)


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

---

## 5) Compressed editorial

- Split each group across tables into chunks of size ≥2. Any such chunking can be represented using only **pairs (2)** and **triples (3)**.
- Let `S = sum(a[i])`. For a fixed table count `T`, total empty seats are `E = T*r - S` (must be ≥0).
- For each group:
  - if `a` even: possible triple counts are `0,2,...,2*(a//6)`
  - if `a` odd: possible triple counts are `1,3,...,1+2*((a-3)//6)`
  Thus:
  - `min_threes` = #odd groups
  - `max_threes` = sum of maximums above
- Binary search minimal `T` from `ceil(S/r)` to `S`.
- `check(T)` tries all total triple counts `threes` in `[min_threes..max_threes]` step 2:
  - require `S-3*threes` is nonnegative and even (rest are pairs).
  - If `r` is odd:
    - if `threes < T`, need `E >= T-threes` (one empty seat for each table without a triple).
    - else put one triple per table: `threes -= T`, `r -= 3` (now even).
  - With even `r`, absorb triples using 6-seat blocks: each 6 seats can hold `3+3` so absorbs 2 triples.
    `absorbed = 2*(r//6)*T`. If remaining triples exceed this, each leftover triple needs one empty seat: require `E >= leftovers`.
- If any `threes` works, `T` is feasible.

This yields the minimal tables.