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;
const int N = 30;
int chd[N], n, need[N];
inline bool cmp(int a, int b) { return a > b; }
void sea(int p) {
  if (p == n) {
    printf("YES\n");
    exit(0);
  }
  need[p] = chd[p] - 1;
  for (int i = 0; chd[i] > chd[p] + 1; i++) {
    if (need[i] >= chd[p]) {
      need[i] -= chd[p];
      sea(p + 1);
      need[i] += chd[p];
    }
  }
  if (chd[p] == 1) {
    printf("NO\n");
    exit(0);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", chd + i);
  }
  sort(chd, chd + n, cmp);
  need[0] = n - 1;
  if (chd[0] == n) sea(1);
  printf("NO\n");
  return 0;
}
