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;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<V<T>>;
template <typename T>
istream &operator>>(istream &cin, V<T> &arr) {
  for (T &val : arr) {
    cin >> val;
  }
  return cin;
}
template <typename T>
ostream &operator<<(ostream &cout, VV<T> &arr) {
  for (V<T> &val : arr) {
    cout << val << endl;
  }
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, V<T> &arr) {
  for (T &val : arr) {
    cout << val << " ";
  }
  return cout;
}
void update(vector<vector<long long>> &tree, long long v, long long l,
            long long r, long long lb, long long rb, long long val) {
  if (lb > rb) {
    return;
  }
  if (l == lb && r == rb) {
    tree[v][2] += val;
  } else {
    long long m = (l + r) / 2;
    update(tree, v * 2, l, m, lb, min(rb, m), val);
    update(tree, v * 2 + 1, m + 1, r, max(lb, m + 1), rb, val);
  }
}
void build(vector<vector<long long>> &tree, vector<long long> &arr, long long v,
           long long l, long long r) {
  if (l == r)
    tree[v] = {arr[l], arr[l]};
  else {
    long long m = (l + r) / 2;
    build(tree, arr, v * 2, l, m);
    build(tree, arr, v * 2 + 1, m + 1, r);
    tree[v][0] = max(tree[v * 2][0], tree[v * 2 + 1][0]);
    tree[v][1] = min(tree[v * 2][1], tree[v * 2 + 1][1]);
  }
}
long long node(long long a, long long b) {
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
void dfs(vector<pair<long long, long long>> &g, vector<bool> &visited, int pos,
         int n) {
  visited[pos] = true;
  for (int i = 0; i < n; i++) {
    if (!visited[i] &&
        (g[pos].first == g[i].first || g[pos].second == g[i].second)) {
      dfs(g, visited, i, n);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<vector<long long>> cntnum(101);
  vector<vector<long long>> cnt(2);
  for (int i = 0; i < n; i++) {
    int num;
    cin >> num;
    cntnum[num].push_back(i);
  }
  for (int i = 1; i <= 100; i++) {
    if (cntnum[i].size() == 1)
      cnt[0].push_back(cntnum[i][0]);
    else if (cntnum[i].size() > 2) {
      for (int j = 0; j < cntnum[i].size(); j++) cnt[1].push_back(cntnum[i][j]);
    }
  }
  vector<string> ans(n);
  for (int i = 0; i < n; i++) ans[i] = 'A';
  int hora = cnt[0].size(), horb = 0;
  for (int i = 0; hora - 1 > horb && i < cnt[0].size(); i++) {
    ans[cnt[0][i]] = 'B';
    hora--;
    horb++;
  }
  for (int i = 0; hora != horb && i < cnt[1].size(); i++) {
    ans[cnt[1][i]] = 'B';
    horb++;
  }
  if (hora != horb)
    cout << "NO";
  else {
    cout << "YES\n";
    for (int i = 0; i < n; i++) cout << ans[i];
  }
  return 0;
}