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 N = 27;
int n, m, a[N], sum[4096], cnt, onecnt;
bool dp[2][4096][N];
int main() {
  scanf("%d", &n);
  for (int i = (1), iE = (n); i <= iE; i++) {
    scanf("%d", &a[i]);
    if (a[i] > 1)
      a[++cnt] = a[i];
    else
      onecnt++;
  }
  if (cnt * 2 + 1 > n) return printf("NO\n"), 0;
  n = cnt, sort(a + 1, a + n + 1);
  for (int i = (1), iE = (n); i <= iE; i++) sum[1 << (i - 1)] = 1;
  dp[0][0][1] = 1;
  for (int i = (2), iE = (onecnt); i <= iE; i++) dp[1][0][i] = 1;
  for (int i = (1), iE = ((1 << n) - 1); i <= iE; i++) {
    sum[i] = ((i & 1) + sum[i >> 1]);
    int t = log2(i), pp = i - (1 << t), ned = a[t + 1] - 1 - sum[pp];
    if (ned >= 0)
      for (int j = (0), jE = (ned); j <= jE; j++) dp[0][i][ned] |= dp[1][pp][j];
    for (int j = (1), jE = (onecnt); j <= jE; j++)
      dp[1][i][j] |= (dp[0][i][j - 1] | dp[1][i][j - 1]);
    for (int k = pp; k; k = ((k - 1) & pp))
      for (int j = (0), jE = (onecnt); j <= jE; j++) {
        for (int d = (0), dE = (j); d <= dE; d++) {
          dp[1][i][j] |= (dp[0][i ^ k][j - d] & (dp[0][k][d] | dp[1][k][d]));
          if (dp[1][i][j]) break;
        }
        if (dp[1][i][j]) break;
      }
  }
  printf("%s\n", dp[0][(1 << n) - 1][onecnt] ? "YES" : "NO");
  return 0;
}
