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 = 24;
int n, nl = 0;
int phia[N];
int* phi = phia;
int A[2048][24];
int B[2048][24];
int pop(int S) {
  int res = 0;
  while (S) S &= S - 1, res++;
  return res;
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &phia[i]);
  sort(phia, phia + n);
  while (n && *phi == 1) phi++, n--, nl++;
  if (nl <= n) {
    printf("NO");
    return 0;
  }
  memset(A, 0, sizeof A);
  memset(B, 0, sizeof B);
  A[0][1] = 1;
  for (int k = 2; k <= nl; k++) B[0][k] = 1;
  for (int S = 1; S < (1 << n); S++) {
    int pS = pop(S);
    for (int x = 0; x < n; x++) {
      if (!(S & (1 << x))) continue;
      int k = phi[x] - pS;
      if (k >= 0 && k <= nl && B[S ^ (1 << x)][k]) A[S][k] = 1;
    }
    for (int S1 = 0; S1 <= S; S1++) {
      if (S1 & (~S)) continue;
      for (int k = 0; k <= nl; k++) {
        for (int k1 = 0; k1 <= k; k1++) {
          if (A[S1][k1] && (A[S ^ S1][k - k1] || B[S ^ S1][k - k1]))
            B[S][k] = 1;
        }
      }
    }
    for (int k = 1; k <= nl; k++)
      if (A[S][k - 1] || B[S][k - 1]) B[S][k] = 1;
  }
  if (A[(1 << n) - 1][nl])
    printf("YES");
  else
    printf("NO");
  return 0;
}
