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, k, d[11], t[11];
vector<int> sz;
bool rec(int x) {
  if (x == n - 1) {
    int m = 0;
    for (auto i = (0); i < n; i++) {
      if (t[i] >= sz[i]) return false;
      int p = sz[i] - 1 - t[i];
      if (d[i] + p < 2) return false;
      m += p;
    }
    return m == k;
  }
  for (auto y = (x + 1); y < n; y++) {
    d[y]++, t[y] += sz[x];
    if (rec(x + 1)) return true;
    d[y]--, t[y] -= sz[x];
  }
  return false;
}
bool work() {
  cin >> n;
  for (auto i = (0); i < n; i++) {
    int s;
    cin >> s;
    if (s == 1)
      k++;
    else if (s == 2)
      return false;
    else
      sz.push_back(s);
  }
  if (2 * k < n) return false;
  sort(begin(sz), end(sz));
  n -= k;
  if (!n) return k == 1;
  return rec(0);
}
int main() {
  cout << (work() ? "YES" : "NO") << endl;
  return 0;
}
