#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const int MOD2 = 998244353;

int k[100010][2];
int a[3];
char ans[100010];

int main(){
    int n;
    cin >> n >> a[0] >> a[1] >> a[2];

    rep(i, n) {
        string s;
        cin >> s;
        k[i][0] = s[0] - 'A';
        k[i][1] = s[1] - 'A';
    }

    rep(i, n){
        if (a[k[i][0]] > a[k[i][1]]) swap(k[i][0], k[i][1]);
        else if (a[k[i][0]] == a[k[i][1]]){
            if (i < n-1 && k[i][1] == k[i+1][0] || k[i][1] == k[i+1][1]){
                swap(k[i][0], k[i][1]);
            }
        }

        if (a[k[i][1]] == 0) {
            cout << "No" << endl;
            return 0;
        }

        a[k[i][0]]++;
        a[k[i][1]]--;
        ans[i] = k[i][0] + 'A';
    }

    cout << "Yes" << endl;
    rep(i, n) cout << ans[i] << endl;

    return 0;
}

### Prompt

Your task is to create 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>
#define rep(i, n) for (int i = 0; i < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const int MOD2 = 998244353;

int k[100010][2];
int a[3];
char ans[100010];

int main(){
    int n;
    cin >> n >> a[0] >> a[1] >> a[2];

    rep(i, n) {
        string s;
        cin >> s;
        k[i][0] = s[0] - 'A';
        k[i][1] = s[1] - 'A';
    }

    rep(i, n){
        if (a[k[i][0]] > a[k[i][1]]) swap(k[i][0], k[i][1]);
        else if (a[k[i][0]] == a[k[i][1]]){
            if (i < n-1 && k[i][1] == k[i+1][0] || k[i][1] == k[i+1][1]){
                swap(k[i][0], k[i][1]);
            }
        }

        if (a[k[i][1]] == 0) {
            cout << "No" << endl;
            return 0;
        }

        a[k[i][0]]++;
        a[k[i][1]]--;
        ans[i] = k[i][0] + 'A';
    }

    cout << "Yes" << endl;
    rep(i, n) cout << ans[i] << endl;

    return 0;
}

```