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>
using namespace std;
int n, c[30];
int f[30][1 << 11][30];
int S[1 << 11];
int cnt = 0, cnt2 = 0;
void dfs(int i, int dep, int j, int l) {
  if (dep > n - cnt) {
    for (int k = 0; k <= cnt; k++)
      if (f[i - 1][j][k]) {
        int dt = c[i] - 1 - S[l];
        if (dt <= k && dt >= 0 && dt + __builtin_popcount(l) >= 2)
          f[i][j | l][k - dt] = 1;
      }
  } else if (dep == i)
    dfs(i, dep + 1, j | (1 << dep - 1), l);
  else {
    dfs(i, dep + 1, j, l);
    dfs(i, dep + 1, j | (1 << dep - 1), l);
    dfs(i, dep + 1, j, l | (1 << dep - 1));
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d", &c[i]), cnt += (c[i] == 1), cnt2 += (c[i] == 2);
  if (n == 1) {
    puts(cnt ? "YES" : "NO");
    return 0;
  }
  if (cnt2 || cnt * 2 <= n) {
    puts("NO");
    return 0;
  }
  sort(c + 1, c + n + 1, greater<int>());
  if (c[1] != n) {
    puts("NO");
    return 0;
  }
  for (int i = 0; i < (1 << n - cnt); i++)
    for (int j = 1; j <= n - cnt; j++)
      if (i & (1 << j - 1)) S[i] += c[j];
  f[0][1][cnt] = 1;
  for (int i = 1; i <= n - cnt; i++) dfs(i, 1, 0, 0);
  puts(f[n - cnt][(1 << n - cnt) - 1][0] ? "YES" : "NO");
  return 0;
}
