Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e12 + 5;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x * f;
}
inline void write(int x) {
  char c[20];
  int l = 0;
  while (x) c[++l] = x % 10 + 48, x /= 10;
  while (l) putchar(c[l--]);
  putchar('\n');
}
long long n, a[105], b[105];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]]++;
  }
  int s0 = 0, num = 0;
  for (int i = 1; i <= 100; ++i) {
    if (b[i] == 1) s0++;
    if (b[i] >= 3) num++;
  }
  if (s0 % 2 == 1 && num == 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  if (s0 % 2 == 0) {
    s0 /= 2;
    for (int i = 1; i <= n; ++i) {
      if (b[a[i]] == 1 && s0) {
        --s0;
        printf("A");
      } else
        printf("B");
    }
  } else {
    bool flag = 0;
    s0 /= 2;
    for (int i = 1; i <= n; ++i) {
      if (b[a[i]] == 1) {
        if (s0) {
          --s0;
          printf("A");
        } else
          printf("B");
      } else if (b[a[i]] >= 3) {
        if (flag == 0) {
          flag = 1;
          printf("A");
        } else
          printf("B");
      } else
        printf("B");
    }
  }
}