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;
struct st {
  int v;
  bool mark;
} a[555];
bool cmp(st a, st b) { return a.v < b.v; }
int n;
void idar(int x, int p, int s, int d) {
  if (x == n) {
    int i, rm = 0;
    for (i = 0; i < n; i++)
      if (a[i].mark == true) rm++;
    if (rm == n - 1) {
      printf("YES\n");
      exit(0);
    }
  }
  if (s == a[x].v - 1 && (d >= 2 || d == 0)) {
    idar(x + 1, 0, 0, 0);
  } else {
    if (p == n) return;
    if (a[p].v + s >= a[x].v) return;
    if (!a[p].mark) {
      a[p].mark = true;
      idar(x, p + 1, s + a[p].v, d + 1);
      a[p].mark = false;
    }
    idar(x, p + 1, s, d);
  }
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i].v);
    a[i].mark = false;
  }
  sort(a, a + n, cmp);
  idar(0, 0, 0, 0);
  printf("NO\n");
  return 0;
}
