#include <bits/stdc++.h>
using namespace std;
int n, a[24], c[25], tag;
char b[1 << 24];
void init() {
  int i;
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  tag = 0;
  for (i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + i);
  if (n == 1 && a[0] != 1 || n == 2 || n > 2 && a[1] != 1) tag = 1;
  for (i = 0; i < n; i++) {
    if (n != 2 && a[i] == 2 || a[i] > n) tag = 1;
  }
  for (i = 0; i < n && a[i] != n; i++)
    ;
  if (i == n) tag = 1;
}
int calc(int u, int al) {
  int i, j, k, l, v;
  c[0] = u;
  for (i = 0; i < n; i++) {
    if ((u & (1 << i)) == 0) continue;
    j = a[i];
    for (k = 24; k - j > -1; k--)
      if (c[k - j] == u) c[k] = u;
  }
  if (c[al] != u) return -1;
  for (i = n - 1, v = u; i > -1; i--) {
    if (u & (1 << i)) {
      if (al >= a[i] && c[al - a[i]] == u) {
        if (al - a[i] == 0) {
          if (v == u)
            continue;
          else
            return v ^ (1 << i);
        }
        al -= a[i];
        v ^= 1 << i;
      }
    }
  }
  return -1;
}
void solve() {
  if (tag) {
    puts("NO");
    return;
  }
  tag = 1 << (n - 1);
  int i, j, k, l;
  int u, v, w;
  queue<int> Q;
  for (i = 0; i < n && a[i] == 1; i++)
    ;
  j = (1 << i) - 1;
  b[j] = 1;
  Q.push(j);
  while (!Q.empty() && !b[tag]) {
    i = Q.front();
    Q.pop();
    for (l = n - 1; (i & (1 << l)) == 0; l--)
      ;
    l++;
    k = calc(i, a[l] - 1);
    if (k != -1) {
      k ^= 1 << l;
      if (!b[k]) {
        b[k] = 1;
        Q.push(k);
      }
    }
  }
  puts(b[tag] ? "YES" : "NO");
}
int main() {
  while (~scanf("%d", &n)) {
    init();
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n, a[24], c[25], tag;
char b[1 << 24];
void init() {
  int i;
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  tag = 0;
  for (i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + i);
  if (n == 1 && a[0] != 1 || n == 2 || n > 2 && a[1] != 1) tag = 1;
  for (i = 0; i < n; i++) {
    if (n != 2 && a[i] == 2 || a[i] > n) tag = 1;
  }
  for (i = 0; i < n && a[i] != n; i++)
    ;
  if (i == n) tag = 1;
}
int calc(int u, int al) {
  int i, j, k, l, v;
  c[0] = u;
  for (i = 0; i < n; i++) {
    if ((u & (1 << i)) == 0) continue;
    j = a[i];
    for (k = 24; k - j > -1; k--)
      if (c[k - j] == u) c[k] = u;
  }
  if (c[al] != u) return -1;
  for (i = n - 1, v = u; i > -1; i--) {
    if (u & (1 << i)) {
      if (al >= a[i] && c[al - a[i]] == u) {
        if (al - a[i] == 0) {
          if (v == u)
            continue;
          else
            return v ^ (1 << i);
        }
        al -= a[i];
        v ^= 1 << i;
      }
    }
  }
  return -1;
}
void solve() {
  if (tag) {
    puts("NO");
    return;
  }
  tag = 1 << (n - 1);
  int i, j, k, l;
  int u, v, w;
  queue<int> Q;
  for (i = 0; i < n && a[i] == 1; i++)
    ;
  j = (1 << i) - 1;
  b[j] = 1;
  Q.push(j);
  while (!Q.empty() && !b[tag]) {
    i = Q.front();
    Q.pop();
    for (l = n - 1; (i & (1 << l)) == 0; l--)
      ;
    l++;
    k = calc(i, a[l] - 1);
    if (k != -1) {
      k ^= 1 << l;
      if (!b[k]) {
        b[k] = 1;
        Q.push(k);
      }
    }
  }
  puts(b[tag] ? "YES" : "NO");
}
int main() {
  while (~scanf("%d", &n)) {
    init();
    solve();
  }
  return 0;
}

```