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 maxn = 30;
int rem[maxn], a[maxn], cnt, n, sum;
bool dfs(int x) {
  if (x < cnt) {
    return sum == cnt;
  }
  for (int i = n - 1; i > x; i--) {
    if (rem[i] > a[x] && a[i] != a[x] + 1) {
      rem[i] -= a[x];
      sum -= a[x];
      if (dfs(x - 1)) return 1;
      sum += a[x];
      rem[i] += a[x];
    }
  }
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (a[i] == 2) {
      puts("NO");
      return 0;
    }
    rem[i] = a[i];
    if (a[i] == 1)
      cnt = i + 1;
    else
      sum += a[i] - 1;
  }
  if (cnt < (n + 1) / 2) {
    puts("NO");
    return 0;
  }
  if (n == 1 && a[0] == 1) {
    puts("YES");
    return 0;
  } else if (n == 1) {
    puts("NO");
    return 0;
  }
  if (dfs(n - 2))
    puts("YES");
  else
    puts("NO");
  return 0;
}
