Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food. 

Iahub asks Iahubina: can you build a rooted tree, such that

  * each internal node (a node with at least one son) has at least two sons; 
  * node i has ci nodes in its subtree? 



Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.

Input

The first line of the input contains integer n (1 ≤ n ≤ 24). Next line contains n positive integers: the i-th number represents ci (1 ≤ ci ≤ n).

Output

Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes). 

Examples

Input

4
1 1 1 4


Output

YES

Input

5
1 1 5 2 1


Output

NO
#include <bits/stdc++.h>
int const maxn = 27;
bool dp[maxn][maxn][1 << (maxn / 2)];
int n, ni, leaf;
std::vector<int> v;
int main() {
  std::cin >> n;
  v.resize(n);
  for (int i = 0; i < n; i++) std::cin >> v[i];
  if (n == 1) {
    std::cout << "YES\n";
    return 0;
  }
  std::sort(v.begin(), v.end());
  int leaf = 0;
  for (; leaf < n; leaf++)
    if (v[leaf] != 1) break;
  if (leaf < (n + 1) / 2 || v[n - 1] != n) {
    std::cout << "NO\n";
    return 0;
  }
  int inter = n - leaf;
  dp[0][leaf][0] = true;
  for (int i = 0; i < inter; i++) {
    for (int j = 0; j <= leaf; j++) {
      for (int st = 0; st < (1 << i); st++) {
        if (!dp[i][j][st]) continue;
        for (int tst = st;; tst = st & (tst - 1)) {
          int sum = 0, tot = 0;
          for (int t = 0; t < i; t++)
            if ((tst >> t) & 1) {
              sum += v[leaf + t];
              tot++;
            }
          int need = v[leaf + i] - sum - 1;
          if (need <= j && need >= 0 && ((!(tot + need) || tot + need > 1)))
            dp[i + 1][j - need][(st ^ tst) | (1 << i)] = true;
          if (!tst) break;
        }
      }
    }
  }
  std::cout << (dp[inter][0][1 << (inter - 1)] ? "YES" : "NO") << '\n';
}
