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 n, a[30], b[30];
bool solve(vector<int> u, int k, int id) {
  if (id == n + 1) {
    if (u.size() != 1) return 0;
    if (k != 0) return 0;
    if (u[0] != n) return 0;
    return 1;
  }
  int m = u.size();
  vector<int> sum(1 << m);
  sum[0] = 0;
  for (int i = 1; i < (1 << m); i++)
    sum[i] = sum[i & (i - 1)] + u[__builtin_ctz(i)];
  for (int i = 0; i < (1 << m); i++)
    if (sum[i] <= (a[id] - 1) && (a[id] - 1) <= sum[i] + k) {
      int r = (a[id] - 1) - sum[i];
      if (r + __builtin_popcount(i) < 2) continue;
      vector<int> v;
      for (int j = 0; j < m; j++)
        if (i >> j & 1 ^ 1) v.push_back(u[j]);
      v.push_back(a[id]);
      if (solve(v, k - r, id + 1) == 1) return 1;
    }
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  if (n == 1) return puts("YES"), 0;
  if (a[n] != n) return puts("NO"), 0;
  int k = 1;
  while (k <= n && a[k] == 1) k++;
  vector<int> u;
  if (solve(u, k - 1, k))
    puts("YES");
  else
    puts("NO");
}
