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;
long long int N;
long long int hashCnt[105];
long long int cnt = 0;
long long int arr[105];
string ans = "";
vector<int> V;
int main() {
  scanf("%lld", &N);
  for (int i = 0; i < N; i++) {
    scanf("%lld", &arr[i]);
    hashCnt[arr[i]]++;
  }
  for (int i = 0; i <= 100; i++) {
    if (hashCnt[i] == 1) {
      cnt++;
      V.push_back(i);
    }
  }
  if (V.size() % 2 == 0) {
    int flag = 0;
    printf("YES\n");
    for (int j = 0; j < N; j++) {
      for (int i = 0; i < V.size(); i++) {
        if (arr[j] == V[i] && flag == 0) {
          cout << "A";
          flag = 1;
        } else if (arr[j] == V[i] && flag == 1) {
          cout << "B";
          flag = 0;
        }
      }
      if (hashCnt[arr[j]] > 1) {
        cout << "A";
      }
    }
  } else {
    int cntA = 0;
    int cntB = 0;
    int flag = 0;
    for (int j = 0; j < N; j++) {
      for (int i = 0; i < V.size(); i++) {
        if (arr[j] == V[i] && flag == 0) {
          ans.push_back('A');
          flag = 1;
        } else if (arr[j] == V[i] && flag == 1) {
          ans.push_back('B');
          flag = 0;
        }
      }
      if (hashCnt[arr[j]] > 1) {
        ans.push_back('A');
      }
    }
    for (int j = 0; j < N; j++) {
      int flag1 = 0;
      for (int i = 0; i < V.size(); i++) {
        if (hashCnt[arr[j]] > 2) {
          ans[j] = 'B';
          flag1 = 1;
          V.push_back(arr[j]);
          break;
        }
      }
      if (flag1) {
        break;
      }
    }
    for (int i = 0; i < ans.size(); i++) {
      if (ans[i] == 'B') {
        cntB++;
      } else {
        cntA++;
      }
    }
    if (V.size() % 2 == 0) {
      printf("YES\n");
      cout << ans;
    } else {
      printf("NO\n");
    }
  }
  return 0;
}