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;
inline long long int gcd(long long int a, long long int b) {
  long long int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
using namespace std;
const int MAXN = (long long int)1e6 + 100;
long long int n, ti;
long long int a[MAXN];
long long int b[MAXN];
bool flag = false;
void DFS(long long int x) {
  ti += 1;
  if (ti >= (long long int)3e7) return;
  if (b[x] + 1 != a[x]) return;
  if (x == n) {
    flag = 1;
    return;
  }
  for (long long int i = x + 1; i <= n; i += 1) {
    if ((a[x] != a[i] - 1) && (b[i] + a[x] <= a[i] - 1)) {
      b[i] += a[x];
      DFS(x + 1);
      if (flag) return;
      b[i] -= a[x];
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long int i = 1; i <= n; i += 1) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  DFS(1);
  if (flag)
    cout << "YES" << endl;
  else
    cout << "NO";
  return 0;
}
