#include <bits/stdc++.h>
using namespace std;
int fn(int curr, int i, int accum, int taken);
int with_me[100], msk[100];
int left1[100], n_left, n_with = 0;
int n, f = 0, x;
int cnt[100] = {0};
int main() {
  for (int i = 0; i < 100; i++) msk[i] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &left1[i]), f += (left1[i] == 2);
  sort(left1, left1 + n);
  int i = 0;
  for (i = 0; i < n && left1[i] == 1; i++) cnt[1]++;
  n_left = i;
  int ans = 0;
  if (f == 0) ans = fn(n_left, 0, 0, 0);
  printf("%s", ans == 1 && !f ? "YES" : "NO");
  return 0;
}
int fn(int curr, int i, int accum, int taken) {
  if (curr == n) {
    int tmp = 0;
    for (int i = 0; i < 100; i++) tmp += cnt[i];
    return tmp == 1;
  }
  if (accum == left1[curr] - 1) {
    if (taken > 1) {
      cnt[left1[curr]]++;
      return fn(curr + 1, 1, 0, 0);
    }
    return 0;
  }
  if (accum > left1[curr] - 1 || i >= 100) return 0;
  int ret = 0;
  int tmp = cnt[i];
  for (int j = 0; j <= tmp && !ret && accum < left1[curr]; j++) {
    cnt[i] -= j;
    ret = ret | fn(curr, i + 1, accum + i * j, taken + j);
    cnt[i] += j;
  }
  return ret;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int fn(int curr, int i, int accum, int taken);
int with_me[100], msk[100];
int left1[100], n_left, n_with = 0;
int n, f = 0, x;
int cnt[100] = {0};
int main() {
  for (int i = 0; i < 100; i++) msk[i] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &left1[i]), f += (left1[i] == 2);
  sort(left1, left1 + n);
  int i = 0;
  for (i = 0; i < n && left1[i] == 1; i++) cnt[1]++;
  n_left = i;
  int ans = 0;
  if (f == 0) ans = fn(n_left, 0, 0, 0);
  printf("%s", ans == 1 && !f ? "YES" : "NO");
  return 0;
}
int fn(int curr, int i, int accum, int taken) {
  if (curr == n) {
    int tmp = 0;
    for (int i = 0; i < 100; i++) tmp += cnt[i];
    return tmp == 1;
  }
  if (accum == left1[curr] - 1) {
    if (taken > 1) {
      cnt[left1[curr]]++;
      return fn(curr + 1, 1, 0, 0);
    }
    return 0;
  }
  if (accum > left1[curr] - 1 || i >= 100) return 0;
  int ret = 0;
  int tmp = cnt[i];
  for (int j = 0; j <= tmp && !ret && accum < left1[curr]; j++) {
    cnt[i] -= j;
    ret = ret | fn(curr, i + 1, accum + i * j, taken + j);
    cnt[i] += j;
  }
  return ret;
}

```