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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
const int N=2e5 + 5;
const ll mod = 1e9 + 7;

int n;
string str[N];
char res[N];

void dfs(int i, int a, int b, int c) {
    if(i > n) {
        cout << "Yes" << endl;
        for(int idx=1;idx<=n;idx++) {
            cout << res[idx] << endl;
        }
        exit(0);
    }
    if(str[i] == "AB") {
        if(a) res[i] = 'B', dfs(i+1, a-1, b+1, c);
        if(b) res[i] = 'A', dfs(i+1, a+1, b-1, c);
    }
    else if(str[i] == "AC") {
        if(a) res[i] = 'C', dfs(i+1, a-1, b, c+1);
        if(c) res[i] = 'A', dfs(i+1, a+1, b, c-1);
    }
    else {
        if(b) res[i] = 'C', dfs(i+1, a, b-1, c+1);
        if(c) res[i] = 'B', dfs(i+1, a, b+1, c-1);
    }
}

int main()
{
    ios_base::sync_with_stdio(0),cin.tie(NULL),cout.tie(NULL);
    int a, b, c;
    cin >> n >> a >> b >> c;
    for(int i=1;i<=n;i++) cin >> str[i];
    dfs(1, a, b, c);
    cout << "No" << endl;
    exit(0);
}