#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

int main() {
    int n;
    int v[3] = {};
    cin >> n >> v[0] >> v[1] >> v[2];
    vector<int> ans;
    bool pos = true;
    vector<string> q(n);
    for (int i = 0; i < n; i++) cin >> q[i];
    for (int i = 0; i < n; i++) {
        int i1 = q[i][0] - 'A', i2 = q[i][1] - 'A';
        if (!v[i1] && !v[i2]) {
            pos = false;
            break;
        }
        if (v[i1] == 1 && v[i2] == 1 && i + 1 < n) {
            int j1 = q[i + 1][0] - 'A', j2 = q[i + 1][1] - 'A';
            if (i1 == j1 || i1 == j2) {
                v[i1]++, v[i2]--;
                ans.push_back(i1);
            } else {
                v[i1]--, v[i2]++;
                ans.push_back(i2);
            }
        } else if (v[i1] > v[i2]) {
            v[i1]--, v[i2]++;
            ans.push_back(i2);
        } else {
            v[i1]++, v[i2]--;
            ans.push_back(i1);
        }
    }
    if (pos) {
        cout << "Yes\n";
        for (auto e : ans) cout << char('A' + e) << "\n";
    } else {
        cout << "No\n";
    }
}

### Prompt

Please formulate a CPP solution to the following problem:

There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice. Each of these choices is represented by a string s_i. If s_i is `AB`, you must add 1 to A or B then subtract 1 from the other; if s_i is `AC`, you must add 1 to A or C then subtract 1 from the other; if s_i is `BC`, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

* 1 \leq N \leq 10^5
* 0 \leq A,B,C \leq 10^9
* N, A, B, C are integers.
* s_i is `AB`, `AC`, or `BC`.

Input

Input is given from Standard Input in the following format:


N A B C
s_1
s_2
:
s_N


Output

If it is possible to make N choices under the condition, print `Yes`; otherwise, print `No`.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (`A`, `B`, or `C`) to which you add 1 in the i-th choice.

Examples

Input

2 1 3 0
AB
AC


Output

Yes
A
C


Input

3 1 0 0
AB
BC
AB


Output

No


Input

1 0 9 0
AC


Output

No


Input

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB


Output

Yes
C
B
B
C
C
B
A
A

### Response

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

using namespace std;
typedef long long ll;

int main() {
    int n;
    int v[3] = {};
    cin >> n >> v[0] >> v[1] >> v[2];
    vector<int> ans;
    bool pos = true;
    vector<string> q(n);
    for (int i = 0; i < n; i++) cin >> q[i];
    for (int i = 0; i < n; i++) {
        int i1 = q[i][0] - 'A', i2 = q[i][1] - 'A';
        if (!v[i1] && !v[i2]) {
            pos = false;
            break;
        }
        if (v[i1] == 1 && v[i2] == 1 && i + 1 < n) {
            int j1 = q[i + 1][0] - 'A', j2 = q[i + 1][1] - 'A';
            if (i1 == j1 || i1 == j2) {
                v[i1]++, v[i2]--;
                ans.push_back(i1);
            } else {
                v[i1]--, v[i2]++;
                ans.push_back(i2);
            }
        } else if (v[i1] > v[i2]) {
            v[i1]--, v[i2]++;
            ans.push_back(i2);
        } else {
            v[i1]++, v[i2]--;
            ans.push_back(i1);
        }
    }
    if (pos) {
        cout << "Yes\n";
        for (auto e : ans) cout << char('A' + e) << "\n";
    } else {
        cout << "No\n";
    }
}

```