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 a[24];
int n;
bool d[25][5000][25];
bool cmp(int a, int b) { return a > b; }
inline bool getbit(int a, int k) { return (a >> k) & 1; }
inline int invbit(int a, int k) { return a ^ (1 << k); }
inline int sum(int m) {
  int r = 0;
  for (int i = 0; i < n; i++) {
    r += getbit(m, i) * a[n - i - 1];
  }
  return r;
}
inline int dbits(int m) {
  int r = 0;
  for (int i = 0; i < n; i++) r += getbit(m, i);
  return r;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1 && a[0] == 1) {
    printf("YES");
    return 0;
  }
  sort(a, a + n, cmp);
  int leafs = 0;
  for (int i = n - 1; i >= 0 && a[i] == 1; i--) leafs++;
  n -= leafs;
  if (n > 11) {
    printf("NO");
    return 0;
  }
  int tmask = 0;
  for (int i = 0; i < n - 1; i++) tmask = invbit(tmask, i);
  d[0][tmask][leafs] = 1;
  for (int i = 0; i < n; i++) {
    for (int m = tmask; m >= 0; m--) {
      for (int l = 0; l < 25; l++) {
        if (!d[i][m][l]) continue;
        for (int s = m;; s = (s - 1) & m) {
          int q = sum(m ^ s);
          int c = dbits(m ^ s);
          if ((a[i] - q - 1) <= l && a[i] >= q && (a[i] - q - 1) + c >= 2) {
            d[i + 1][s][l - (a[i] - q - 1)] = 1;
          }
          if (s == 0) break;
        }
      }
    }
  }
  if (d[n][0][0]) {
    printf("YES");
  } else {
    printf("NO");
  }
  return 0;
}
