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], cnt[30], now[30];
int work(int i, int cur, int ch, int x) {
  if (cur == 1) {
    if (ch < 2) return 0;
    if (i == n) return 1;
    now[c[i]]++;
    return work(i + 1, c[i + 1], 2 * (c[i + 1] == 1), 1);
  }
  if (x >= cur) return 0;
  if (now[x] > 0) {
    now[x]--;
    if (work(i, cur - x, min(2, ch + 1), x)) return 1;
    now[x]++;
  }
  if (work(i, cur, ch, x + 1)) return 1;
  return 0;
}
int main() {
  scanf("%d", &n);
  int lf = n / 2 + 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    cnt[c[i]]++;
  }
  sort(c + 1, c + n + 1);
  if (c[n] == n && work(1, c[1], 2 * (c[1] == 1), 1))
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
