Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

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

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit32(x) (1 << (x))
#define bit64(x) (1ll << (x))
#define sz(v) ((int) v.size())

using namespace std;

using i64 = long long;
using f80 = long double;
using vi32 = vector<int>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << endl; exit(0); }
inline void no() { cout << "No" << endl; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); return a / gcd(a, b) * b; }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { x = max(x, y); }
template <typename T> inline void amin(T &x, T y) { x = min(x, y); }
template <typename T> inline T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { each(x, v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

int dpl[100001][4][4][4];

void solve() {
  int n, a, b, c;
  cin >> n >> a >> b >> c;
  vstr s(n);
  cin >> s;
  amin(a, 2);
  amin(b, 2);
  amin(c, 2);
  dpl[0][a][b][c] = 1;
  rep(i, n) repc(j, 3) repc(k, 3) repc(l, 3) {
    if (dpl[i][j][k][l] == 0) continue;
    if (s[i] == "AB") {
      if (j > 0) dpl[i+1][j-1][k+1][l] = 1;
      if (k > 0) dpl[i+1][j+1][k-1][l] = 1;
    } else if (s[i] == "BC") {
      if (k > 0) dpl[i+1][j][k-1][l+1] = 1;
      if (l > 0) dpl[i+1][j][k+1][l-1] = 1;
    } else {
      if (j > 0) dpl[i+1][j-1][k][l+1] = 1;
      if (l > 0) dpl[i+1][j+1][k][l-1] = 1;
    }
  }
  bool ok = false;
  int nj, nk, nl;
  repc(j, 3) repc(k, 3) repc(l, 3) {
    if (!ok && dpl[n][j][k][l]) {
      ok = true;
      nj = j, nk = k, nl = l;
    }
  }
  if (!ok) {
    cout << "No" << endl;
    return;
  }
  string ans;
  while (sz(ans) < n) {
    int i = n - sz(ans) - 1;
    int j = nj, k = nk, l = nl;
    if (s[i] == "AB") {
      if (j > 0 && dpl[i][j-1][k+1][l]) {
        nj = j-1, nk = k+1, nl = l, ans += 'A';
      } else if (k > 0 && dpl[i][j+1][k-1][l]) {
        nj = j+1, nk = k-1, nl = l, ans += 'B';
      }
    } else if (s[i] == "BC") {
      if (k > 0 && dpl[i][j][k-1][l+1]) {
        nj = j, nk = k-1, nl = l+1, ans += 'B';
      } else if (l > 0 && dpl[i][j][k+1][l-1]) {
        nj = j, nk = k+1, nl = l-1, ans += 'C';
      }
    } else {
      if (j > 0 && dpl[i][j-1][k][l+1]) {
        nj = j-1, nk = k, nl = l+1, ans += 'A';
      } else if (l > 0 && dpl[i][j+1][k][l-1]) {
        nj = j+1, nk = k, nl = l-1, ans += 'C';
      }
    }
  }
  cout << "Yes" << endl;
  reverse(all(ans));
  rep(i, n) {
    cout << ans[i] << endl;
  }
}