#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const bool db = false;
int n, ones;
int c[33], u[33], cc;
int ch[33], hv[33], nd[33];
void NO() {
  puts("NO");
  exit(0);
}
void YES() {
  puts("YES");
  exit(0);
}
void calc(int pos, int pt) {
  if (pos == n + 1) {
    bool ok = true;
    for (int i = 1; i <= n && ok; ++i) ok &= hv[i] == nd[i];
    if (ok) YES();
    return;
  }
  for (; pt >= 0 && !ch[pt]; --pt)
    ;
  hv[pos] = 1;
  nd[pos] = pt;
  bool ok = true;
  for (int i = 1; i <= pos && ok; ++i)
    if (hv[i] + pt <= nd[i] && nd[i] - 1 > pt) {
      hv[i] += pt;
      --ch[pt];
      calc(pos + 1, pt);
      hv[i] -= pt;
      ++ch[pt];
      ok &= pt != 1;
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  sort(c + 1, c + n + 1);
  int nn = 0;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    nn += c[i] == n;
    ones += c[i] == 1;
    if (c[i] != n) v.push_back(c[i]);
    ++ch[c[i]];
  }
  if (n == 1) {
    if (c[1] == 1) YES();
    NO();
  }
  if (nn != 1 || ones * 2 <= n) NO();
  hv[1] = 1;
  nd[1] = n;
  calc(2, n - 1);
  NO();
  getchar();
  getchar();
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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>
#pragma comment(linker, "/STACK:100000000")
using namespace std;
const bool db = false;
int n, ones;
int c[33], u[33], cc;
int ch[33], hv[33], nd[33];
void NO() {
  puts("NO");
  exit(0);
}
void YES() {
  puts("YES");
  exit(0);
}
void calc(int pos, int pt) {
  if (pos == n + 1) {
    bool ok = true;
    for (int i = 1; i <= n && ok; ++i) ok &= hv[i] == nd[i];
    if (ok) YES();
    return;
  }
  for (; pt >= 0 && !ch[pt]; --pt)
    ;
  hv[pos] = 1;
  nd[pos] = pt;
  bool ok = true;
  for (int i = 1; i <= pos && ok; ++i)
    if (hv[i] + pt <= nd[i] && nd[i] - 1 > pt) {
      hv[i] += pt;
      --ch[pt];
      calc(pos + 1, pt);
      hv[i] -= pt;
      ++ch[pt];
      ok &= pt != 1;
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  sort(c + 1, c + n + 1);
  int nn = 0;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    nn += c[i] == n;
    ones += c[i] == 1;
    if (c[i] != n) v.push_back(c[i]);
    ++ch[c[i]];
  }
  if (n == 1) {
    if (c[1] == 1) YES();
    NO();
  }
  if (nn != 1 || ones * 2 <= n) NO();
  hv[1] = 1;
  nd[1] = n;
  calc(2, n - 1);
  NO();
  getchar();
  getchar();
  return 0;
}

```