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 maxN = 26;
const int maxS = (1 << 13) + 5;
void no() {
  cout << "NO" << endl;
  exit(0);
}
void yes() {
  cout << "YES" << endl;
  exit(0);
}
int n, m;
vector<int> c;
int d[maxN][maxN][maxS];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  int leaf = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x == 1)
      leaf++;
    else
      c.push_back(x);
  }
  if (n == 1) {
    if (leaf == 1)
      yes();
    else
      no();
  }
  int m = ((int)(c).size());
  if (leaf * 2 <= n || leaf == n) no();
  sort(c.begin(), c.end());
  for (int i = 0; i < n; i++)
    if (c[i] == 2) no();
  int pw = 0;
  if (m != 0) pw = (1 << (m - 1)) - 1;
  if (leaf < c[0] - 1) no();
  d[0][leaf - (c[0] - 1)][0] = 1;
  for (int i = 1; i < m; i++) {
    for (int msk = 0; msk < (1 << i); msk++)
      for (int j = 0; j <= leaf; j++) {
        if (!d[i - 1][j][msk]) continue;
        for (int x = 0; x < (1 << i); x++) {
          if ((x & msk) != 0) continue;
          int sum = 0;
          for (int k = 0; k < i; k++)
            if (x >> k & 1) sum += c[k];
          if (sum > c[i] - 1 || sum + j < c[i] - 1) continue;
          if (__builtin_popcount(x) + c[i] - 1 - sum == 1) continue;
          d[i][j - (c[i] - 1 - sum)][msk ^ x] = 1;
        }
      }
  }
  if (d[m - 1][0][pw])
    yes();
  else
    no();
  return 0;
}
