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;
set<int> second[30];
int n, a[200];
void dfs(int mask, int t) {
  second[t].insert(mask);
  if (mask == (1 << (n - 1)) - 1) {
    cout << "YES";
    exit(0);
  }
  if (t == n) return;
  int b[30], k = 0, c[30];
  for (int i = 0; i <= 30; i++) c[i] = 0;
  for (int i = 1; i <= t; i++)
    if (mask & (1 << (i - 1)))
      b[i] = 1;
    else {
      c[k] = i;
      k++;
    }
  for (int i = 0; i < (1 << k); i++) {
    int f = 0;
    int s = 0;
    int cc = 0;
    for (int j = 0; j < k; j++)
      if (i & (1 << j)) {
        s += a[c[j]];
        f += (1 << (c[j] - 1));
        cc++;
      }
    if (s + 1 == a[t + 1] && cc > 1 && second[t + 1].count(mask + f) == 0)
      dfs(mask + f, t + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int t = 0;
  while (a[t + 1] == 1) t++;
  dfs(0, t);
  cout << "NO";
}
