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 int N = 2e5 + 5;
vector<pair<int, int>> v;
map<int, int> mp, mp2;
vector<int> vv;
int dp[105][105][105] = {};
int solve(int idx, int sz1, int sz2) {
  if (idx >= (int)v.size()) return ((sz1 == sz2));
  int &ret = dp[idx][sz1][sz2];
  if (ret != -1) return ret;
  ret = 0;
  if (v[idx].second > 1) ret |= solve(idx + 1, sz1, sz2);
  if (v[idx].second == 2)
    ret |= solve(idx + 1, sz1 + 1, sz2 + 1);
  else {
    ret |= solve(idx + 1, sz1 + 1, sz2);
    ret |= solve(idx + 1, sz1, sz2 + 1);
  }
  return ret;
}
unordered_set<int> ans;
void output(int idx, int sz1, int sz2) {
  if (idx >= (int)v.size()) return;
  int &ret = dp[idx][sz1][sz2];
  if (v[idx].second > 1) {
    if (ret == solve(idx + 1, sz1, sz2)) {
      output(idx + 1, sz1, sz2);
      return;
    }
  }
  if (v[idx].second == 2) {
    if (ret == solve(idx + 1, sz1 + 1, sz2 + 1)) {
      mp2[v[idx].first] = 1;
      output(idx + 1, sz1 + 1, sz2 + 1);
    }
  } else {
    if (ret == solve(idx + 1, sz1 + 1, sz2)) {
      mp2[v[idx].first] = 1;
      output(idx + 1, sz1 + 1, sz2);
    } else if (ret == solve(idx + 1, sz1, sz2 + 1)) {
      mp2[v[idx].first] = v[idx].second - 1;
      output(idx + 1, sz1, sz2 + 1);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), ios_base::sync_with_stdio(0), cin.tie(0),
      cout.tie(0);
  ;
  memset(dp, -1, sizeof(dp));
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    mp[x]++;
    vv.push_back(x);
  }
  for (auto u : mp) v.push_back({u.first, u.second});
  if (solve(0, 0, 0)) {
    cout << "YES\n";
    output(0, 0, 0);
    for (auto u : vv) {
      if (mp2[u])
        cout << "A", mp2[u]--;
      else
        cout << "B";
    }
    return 0;
  }
  return cout << "NO"
              << "\n",
         0;
  return 0;
}