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 m;
int dp[5000][30], forest[5000][30], c[30];
void lemon() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%d", &c[i]);
  int cnt = 0;
  for (int i = (1); i <= (n); i++)
    if (c[i] == 1) cnt++;
  if (n - cnt > 12) {
    cout << "NO" << endl;
    return;
  }
  if (n == 1) {
    if (c[1] == 1)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
    return;
  }
  int all = 0;
  for (int i = (1); i <= (n); i++)
    if (c[i] != 1) all++, c[all] = c[i];
  m = all;
  for (int leaf = (1); leaf <= (cnt); leaf++) forest[0][leaf] = 1;
  for (int state = (1); state <= ((1 << m) - 1); state++)
    for (int leaf = (2); leaf <= (cnt); leaf++) {
      int maxv = 0, maxvi = 0;
      for (int i = (1); i <= (m); i++)
        if (state & (1 << (i - 1)))
          if (c[i] > maxv) maxv = c[i], maxvi = i;
      int inode = 0;
      for (int i = (1); i <= (m); i++)
        if (state & (1 << (i - 1))) inode++;
      if (inode + leaf == maxv) {
        int s = state - (1 << (maxvi - 1));
        int cs = s;
        if (s == 0) dp[state][leaf] = 1;
        while (cs) {
          if (dp[state][leaf]) break;
          for (int k = (0); k <= (leaf); k++)
            if (dp[cs][k] && forest[s - cs][leaf - k]) dp[state][leaf] = 1;
          cs = (cs - 1) & s;
        }
      }
      if (dp[state][leaf]) forest[state][leaf] = 1;
      if (!forest[state][leaf])
        for (int ss = (0); ss <= (state); ss++)
          for (int i = (1); i <= (leaf); i++)
            if (forest[ss][i] && forest[state - ss][leaf - i]) {
              forest[state][leaf] = 1;
              break;
            }
    }
  if (dp[(1 << m) - 1][cnt])
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
int main() {
  ios::sync_with_stdio(true);
  lemon();
  return 0;
}
