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 double TL = 0.95;
int nc[26], cur[26], cnt[26];
int n;
int p[26];
clock_t start;
inline bool TLE(void) { return (clock() - start > TL * CLOCKS_PER_SEC); }
void init(void) {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i = i + 1) scanf("%d", &nc[i]);
  sort(nc + 1, nc + n + 1);
  if (n == 1) {
    if (nc[1] == 1)
      printf("YES");
    else
      printf("NO");
    exit(0);
  }
  if (nc[1] > 1) {
    printf("NO");
    exit(0);
  }
  for (int i = (1); i <= (n); i = i + 1) cur[i] = nc[i];
}
bool check(void) {
  for (int i = (1); i <= (n); i = i + 1)
    if (nc[i] != 1) return (false);
  return (true);
}
void backtrack(int x) {
  if (TLE()) {
    printf("NO");
    exit(0);
  }
  if (cur[x] != 1 || cnt[x] == 1) return;
  if (nc[x + 1] > 1 && cnt[x + 1] == 0) return;
  for (int i = (x + 1); i <= (n); i = i + 1)
    if (cur[i] > nc[x])
      if (cur[i] == nc[x] + 1 || (i > x + 1 && cur[i] - nc[x] > nc[x + 1])) {
        p[x] = i;
        cnt[i]++;
        cur[i] -= nc[x];
        if (x == n - 1) {
          if (cur[n] == 1) {
            printf("YES");
            exit(0);
          }
        } else
          backtrack(x + 1);
        cur[i] += nc[x];
        cnt[i]--;
      }
}
void process(void) {
  backtrack(1);
  printf("NO");
}
int main(void) {
  start = clock();
  init();
  process();
}
