#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

void yes() { cout << "Yes" << endl; }

void no() {
  cout << "No" << endl;
  exit(0);
}

vector<ll> a, ans;
ll n;
vector<string> s;

void dfs(ll p) {
  if (p == n) {
    yes();
    REP(i, n) { cout << (char)(ans[i] + 'A') << endl; }
    exit(0);
  }

  REP(i, 2) {
    ll plus = s[p][i] - 'A', minus = s[p][i ^ 1] - 'A';
    if (a[minus] == 0) {
      continue;
    }
    a[plus]++, a[minus]--;
    ans.push_back(plus);
    dfs(p + 1);
    ans.pop_back();
    a[plus]--, a[minus]++;
  }
}

int main() {
  cin >> n;
  a = vector<ll>(3, 0);
  REP(i, 3) { cin >> a[i]; }
  s = vector<string>(n);
  REP(i, n) { cin >> s[i]; }
  dfs(0);
  no();
}
### Prompt

In cpp, your task is to solve 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;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

void yes() { cout << "Yes" << endl; }

void no() {
  cout << "No" << endl;
  exit(0);
}

vector<ll> a, ans;
ll n;
vector<string> s;

void dfs(ll p) {
  if (p == n) {
    yes();
    REP(i, n) { cout << (char)(ans[i] + 'A') << endl; }
    exit(0);
  }

  REP(i, 2) {
    ll plus = s[p][i] - 'A', minus = s[p][i ^ 1] - 'A';
    if (a[minus] == 0) {
      continue;
    }
    a[plus]++, a[minus]--;
    ans.push_back(plus);
    dfs(p + 1);
    ans.pop_back();
    a[plus]--, a[minus]++;
  }
}

int main() {
  cin >> n;
  a = vector<ll>(3, 0);
  REP(i, 3) { cin >> a[i]; }
  s = vector<string>(n);
  REP(i, n) { cin >> s[i]; }
  dfs(0);
  no();
}
```