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 inf = 1e9;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
int n, dp[21][1 << 13][20], sum[1 << 15], a[N], x, one;
bool f(int node, int mask, int ret) {
  if (ret < 0) return false;
  if (node == 0 && !ret) return true;
  if (node == 0) return false;
  int &r = dp[node][mask][ret];
  if (r != -1) return r;
  r = false;
  for (int i = 0; i <= (1 << node - 1) - 1; i++)
    if ((i & mask) == false && sum[i] <= a[node] - 1 &&
        __builtin_popcount(i) + a[node] - sum[i] - 1 >= 2)
      r = max(r, (int)f(node - 1, mask | i, ret - (a[node] - 1 - sum[i])));
  return r;
}
int main() {
  scanf("%d", &n);
  int s = 0;
  memset(dp, -1, sizeof dp);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    if (x == 1)
      one++;
    else
      a[++s] = x;
  }
  if (n == 1 && one == 1) {
    cout << "YES" << '\n';
    return 0;
  }
  if (one < n / 2) {
    cout << "NO" << '\n';
    return 0;
  }
  n = s;
  sort(a + 1, a + n + 1);
  for (int i = 0; i <= (1 << n) - 1; i++)
    for (int j = 1; j <= n; j++)
      if ((1 << j - 1) & i) sum[i] += a[j];
  cout << (f(n, 0, one) ? "YES" : "NO") << '\n';
  return 0;
}
