#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i, n) for(int i = 0; i < n; i++)
#define println(v) cout << v << "\n";

inline vector<vector<ll>> make_vector(ll i, ll j) {
    vector<vector<ll>> v(i, vector<ll>(j, 0));
    return v;
}

int main()
{

    int N;
    cin >> N;
    vector<ll> var(3);
    REP(i, 3) cin >> var[i];

    auto option = make_vector(N, 2);
    REP(i, N) {
        string s;
        cin >> s;
        option[i][0] = s[0]-'A';
        option[i][1] = s[1]-'A';
    }

    vector<ll> ans(N);
    REP(i, N) {
        int i1 = option[i][0];
        int i2 = option[i][1];
        int iM = (var[i1] >= var[i2] ? i1 : i2);
        int im = i1 ^ i2 ^ iM;
        if (var[i1] == 1 && var[i2] == 1 && var[i1^i2^3] == 0 && i+1 < N && (option[i+1][0] == i1 || option[i+1][1] == i1)) {
            swap(iM, im);
        }
        var[iM]--;
        var[im]++;
        ans[i] = im;
        if (var[0] < 0 || var[1] < 0 || var[2] < 0) {
            println("No");
            return 0;
        }
    }

    println("Yes");
    REP(i, N) {
        println((char)('A'+ans[i]));
    }

    return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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;
#define REP(i, n) for(int i = 0; i < n; i++)
#define println(v) cout << v << "\n";

inline vector<vector<ll>> make_vector(ll i, ll j) {
    vector<vector<ll>> v(i, vector<ll>(j, 0));
    return v;
}

int main()
{

    int N;
    cin >> N;
    vector<ll> var(3);
    REP(i, 3) cin >> var[i];

    auto option = make_vector(N, 2);
    REP(i, N) {
        string s;
        cin >> s;
        option[i][0] = s[0]-'A';
        option[i][1] = s[1]-'A';
    }

    vector<ll> ans(N);
    REP(i, N) {
        int i1 = option[i][0];
        int i2 = option[i][1];
        int iM = (var[i1] >= var[i2] ? i1 : i2);
        int im = i1 ^ i2 ^ iM;
        if (var[i1] == 1 && var[i2] == 1 && var[i1^i2^3] == 0 && i+1 < N && (option[i+1][0] == i1 || option[i+1][1] == i1)) {
            swap(iM, im);
        }
        var[iM]--;
        var[im]++;
        ans[i] = im;
        if (var[0] < 0 || var[1] < 0 || var[2] < 0) {
            println("No");
            return 0;
        }
    }

    println("Yes");
    REP(i, N) {
        println((char)('A'+ans[i]));
    }

    return 0;
}
```