Vasya has a multiset s consisting of n integer numbers. Vasya calls some number x nice if it appears in the multiset exactly once. For example, multiset \{1, 1, 2, 3, 3, 3, 4\} contains nice numbers 2 and 4.

Vasya wants to split multiset s into two multisets a and b (one of which may be empty) in such a way that the quantity of nice numbers in multiset a would be the same as the quantity of nice numbers in multiset b (the quantity of numbers to appear exactly once in multiset a and the quantity of numbers to appear exactly once in multiset b).

Input

The first line contains a single integer n~(2 ≤ n ≤ 100).

The second line contains n integers s_1, s_2, ... s_n~(1 ≤ s_i ≤ 100) — the multiset s.

Output

If there exists no split of s to satisfy the given requirements, then print "NO" in the first line.

Otherwise print "YES" in the first line.

The second line should contain a string, consisting of n characters. i-th character should be equal to 'A' if the i-th element of multiset s goes to multiset a and 'B' if if the i-th element of multiset s goes to multiset b. Elements are numbered from 1 to n in the order they are given in the input.

If there exist multiple solutions, then print any of them.

Examples

Input

4
3 5 7 1


Output

YES
BABA


Input

3
3 5 1


Output

NO
#include <bits/stdc++.h>
using namespace std;
const long long Maxn = 1e5 + 7;
const long long Inf = 1e9 + 7;
int N;
int vis[110];
struct node {
  int x, num, id;
  char cc;
} A[107];
bool cmp(node a, node b) {
  if (a.num != b.num) return a.num < b.num;
  return a.x < b.x;
}
bool cmp1(node a, node b) { return a.id < b.id; }
int main() {
  while (~scanf(" %d", &N)) {
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= N; i++) {
      scanf(" %d", &A[i].x);
      A[i].id = i;
      vis[A[i].x]++;
    }
    for (int i = 1; i <= N; i++) {
      A[i].num = vis[A[i].x];
    }
    sort(A + 1, A + N + 1, cmp);
    int ct1 = 0, ct2 = 0;
    for (int i = 1; i <= N; i++) {
      if (A[i].num == 1) {
        if (ct1 <= ct2) {
          A[i].cc = 'A';
          ct1++;
        } else {
          A[i].cc = 'B';
          ct2++;
        }
      } else if (A[i].num == 2) {
        if (A[i].x != A[i - 1].x) {
          A[i].cc = 'A';
          ct1++;
        } else {
          A[i].cc = 'B';
          ct2++;
        }
      } else if (A[i].num > 2) {
        if (ct1 == ct2) {
          A[i].cc = 'A';
        } else if (ct1 != ct2) {
          ct2++;
          A[i].cc = 'B';
        }
      }
    }
    if (ct1 == ct2) {
      printf("YES\n");
    } else {
      printf("NO\n");
      continue;
    }
    sort(A + 1, A + N + 1, cmp1);
    for (int i = 1; i <= N; i++) {
      printf("%c", A[i].cc);
    }
    printf("\n");
  }
  return 0;
}
