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 int S = 123;
int main() {
  int n;
  while (cin >> n) {
    int ara[105];
    int inp[n + 5];
    memset(ara, 0, sizeof(ara));
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      inp[i] = x;
      ara[x]++;
    }
    int one = 0, two = 0;
    for (int i = 1; i <= 100; i++) {
      if (ara[i] == 1)
        one++;
      else if (ara[i] == 2)
        two++;
    }
    if (one == 0 && two == 0) {
      puts("YES");
      for (int i = 0; i < n; i++) printf("A");
      printf("\n");
    } else if (one % 2 == 0) {
      puts("YES");
      char ans[n + 5];
      ans[0] = '\0';
      for (int i = 0; i < n; i++) {
        int temp = inp[i];
        if (ara[temp] >= 3) ans[i] = 'A';
      }
      int ro = one / 2;
      int cnt = 1;
      for (int i = 0; i < n; i++) {
        int temp = inp[i];
        if (ara[temp] == 1) {
          if (cnt <= ro)
            ans[i] = 'A', cnt++;
          else if (cnt > ro)
            ans[i] = 'B', cnt++;
          ara[temp] = 0;
        }
      }
      int rt = two;
      for (int i = 0; i < n; i++) {
        int temp = inp[i];
        if (ara[temp] == 2) {
          ans[i] = 'A';
          ara[temp]--;
        } else if (ara[temp] == 1) {
          ans[i] = 'B';
          ara[temp]--;
        }
      }
      ans[n] = '\0';
      puts(ans);
    } else if (one % 2 == 1) {
      bool flag = false;
      int chor, ind;
      for (int i = 0; i < n; i++) {
        int temp = inp[i];
        if (ara[temp] >= 3) {
          flag = true;
          break;
        }
      }
      if (flag) {
        puts("YES");
        char ans[n + 5];
        ans[0] = '\0';
        int a = 0, b = 0;
        int ro = (one + 1) / 2;
        int cnt = 1;
        for (int i = 0; i < n; i++) {
          int temp = inp[i];
          if (ara[temp] == 1) {
            if (cnt <= ro) {
              ans[i] = 'A', a++;
              cnt++;
            } else if (cnt > ro)
              ans[i] = 'B', b++, cnt++;
            ara[temp] = 0;
          }
        }
        int rt = two;
        for (int i = 0; i < n; i++) {
          int temp = inp[i];
          if (ara[temp] == 2) {
            ans[i] = 'A';
            ara[temp]--;
          } else if (ara[temp] == 1) {
            ans[i] = 'B';
            ara[temp]--;
          }
        }
        for (int i = 0; i < n; i++) {
          int temp = inp[i];
          if (ara[temp] >= 3) {
            if (a > b) {
              ans[i] = 'B';
              for (int j = i + 1; j < n; j++) {
                int x = inp[j];
                if (ara[x] >= 3) ans[j] = 'A';
              }
              break;
            } else {
              ans[i] = 'A';
              for (int j = i + 1; j < n; j++) {
                int x = inp[j];
                if (ara[x] >= 3) ans[j] = 'B';
              }
              break;
            }
          }
        }
        ans[n] = '\0';
        puts(ans);
      } else
        puts("NO");
    } else
      puts("NO");
  }
  return 0;
}
