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 a[30];
bool b[30];
int n;
bool go(int s, int i, int j) {
  if (j < 0) return false;
  if (a[j] >= a[i] - 1 || b[j] || s + a[j] > a[i]) return go(s, i, j - 1);
  b[j] = true;
  s += a[j];
  if (s == a[i]) return true;
  if (go(s, i, j - 1))
    return true;
  else {
    b[j] = false;
    return false;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  bool ok = true;
  memset(b, false, sizeof(b));
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) continue;
    int s = 1;
    ok = false;
    for (int j = i - 1; j >= 0; j--) {
      if (go(s, i, j)) {
        ok = true;
        break;
      }
    }
    if (!ok) break;
  }
  if (a[n - 1] != n) ok = false;
  if (ok)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
