#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int maxn = 100010;
int c[30], vis[30], s[30][30], n;
bool cmp(int a, int b) { return a > b; }
bool dfs1(int);
bool dfs2(int p, int ip, int val) {
  int i;
  if (val == 0) {
    if (s[p][0] < 2) return false;
    for (i = 1; i <= s[p][0]; i++)
      if (!dfs1(s[p][i])) return false;
    return true;
  }
  if (ip >= n) return false;
  for (i = ip; i < n; i++)
    if (!vis[i] && val >= c[i]) {
      vis[i] = 1;
      s[p][++s[p][0]] = i;
      if (dfs2(p, i + 1, val - c[i])) return true;
      s[p][0]--;
      vis[i] = 0;
    }
  return false;
}
bool dfs1(int x) {
  vis[x] = 1;
  if (c[x] == 1) return true;
  if (dfs2(x, x + 1, c[x] - 1))
    return true;
  else {
    vis[x] = 0;
    return false;
  }
}
int main() {
  int i, flag, ct;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", &c[i]);
    sort(c, c + n, cmp);
    memset(vis, 0, sizeof vis);
    for (i = 0; i < n; i++) s[i][0] = 0;
    flag = dfs1(0), ct = 0;
    for (i = 0; i < n; i++) ct += vis[i];
    if (ct == n && flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### 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;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int maxn = 100010;
int c[30], vis[30], s[30][30], n;
bool cmp(int a, int b) { return a > b; }
bool dfs1(int);
bool dfs2(int p, int ip, int val) {
  int i;
  if (val == 0) {
    if (s[p][0] < 2) return false;
    for (i = 1; i <= s[p][0]; i++)
      if (!dfs1(s[p][i])) return false;
    return true;
  }
  if (ip >= n) return false;
  for (i = ip; i < n; i++)
    if (!vis[i] && val >= c[i]) {
      vis[i] = 1;
      s[p][++s[p][0]] = i;
      if (dfs2(p, i + 1, val - c[i])) return true;
      s[p][0]--;
      vis[i] = 0;
    }
  return false;
}
bool dfs1(int x) {
  vis[x] = 1;
  if (c[x] == 1) return true;
  if (dfs2(x, x + 1, c[x] - 1))
    return true;
  else {
    vis[x] = 0;
    return false;
  }
}
int main() {
  int i, flag, ct;
  while (~scanf("%d", &n)) {
    for (i = 0; i < n; i++) scanf("%d", &c[i]);
    sort(c, c + n, cmp);
    memset(vis, 0, sizeof vis);
    for (i = 0; i < n; i++) s[i][0] = 0;
    flag = dfs1(0), ct = 0;
    for (i = 0; i < n; i++) ct += vis[i];
    if (ct == n && flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```