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 c[30];
int dp[(1 << 12) + 3][13][25];
int sum[1 << 12];
int cnt[1 << 12];
int tot;
int r(int mask, int p, int child) {
  if (c[p] == 2) return 0;
  if (child < 0) return 0;
  if ((mask >> p) & 1) return 0;
  if (p == tot) {
    return (child == 0);
  }
  if (dp[mask][p][child] != -1) return dp[mask][p][child];
  int res = r(mask, p + 1, child - c[p] + 1);
  if (res) return dp[mask][p][child] = 1;
  for (int s = mask; s > 0; s = ((s - 1) & mask)) {
    int k = cnt[s];
    int val = sum[s] + 1;
    if (val > c[p]) continue;
    k += c[p] - sum[s] - 1;
    if (k < 2) continue;
    int res = r(mask ^ s, p + 1, child - (c[p] - sum[s] - 1));
    if (res) return dp[mask][p][child] = 1;
  }
  return dp[mask][p][child] = 0;
}
int main() {
  int n;
  scanf("%d", &n);
  int v = 0;
  for (int i = (0); i < (n); ++i) {
    scanf("%d", &c[i]);
    if (c[i] == 1) v++;
  }
  if (v < (n + 1) / 2) {
    cout << "NO" << endl;
    return 0;
  }
  if (n == 1) {
    cout << "YES" << endl;
    return 0;
  }
  sort(c, c + n);
  reverse(c, c + n);
  for (int i = (0); i < (n); ++i) {
    if (c[i] > 1) tot++;
  }
  for (int i = (0); i < ((1 << tot)); ++i) {
    for (int j = (0); j < (tot); ++j) {
      if ((i >> j) & 1) {
        sum[i] += c[j];
        cnt[i]++;
      }
    }
  }
  memset(dp, -1, sizeof(dp));
  int res = r((1 << tot) - 2, 0, v);
  if (res)
    printf("YES\n");
  else
    printf("NO\n");
  return 0;
}
