#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int c[30];
int n, leaf, tot_sum, pos;
using pii = pair<int, int>;
void nof() {
  puts("NO");
  exit(0);
}
void yesf() {
  puts("YES");
  exit(0);
}
vector<pii> vec;
const double TL = 0.85;
void dfs(int k) {
  if (k == pos) {
    yesf();
    return;
  }
  vector<pii> V = vec;
  sort(V.begin(), V.end());
  if (clock() * 1.0 / CLOCKS_PER_SEC > TL) return;
  for (int i = 0; i < V.size(); i++) {
    if (i > 0 && V[i] == V[i - 1]) continue;
    if (clock() * 1.0 / CLOCKS_PER_SEC > TL) break;
    pii x = V[i];
    if (x.first < c[k]) continue;
    pii e, A = pii(-1, -1), B = pii(-1, -1);
    if (x.first == c[k] && x.second == 2) continue;
    for (auto &y : vec) {
      if (y == x) {
        swap(vec.back(), y);
        vec.pop_back();
        break;
      }
    }
    e = x;
    x.first -= c[k];
    x.second = 1;
    if (x.first) {
      vec.push_back(x);
      A = x;
    }
    B = pii(c[k] - 1, 2);
    vec.push_back(B);
    dfs(k + 1);
    if (A.first > 0) {
      for (auto &y : vec) {
        if (y == A) {
          swap(y, vec.back());
          vec.pop_back();
          break;
        }
      }
    }
    for (auto &y : vec) {
      if (y == B) {
        swap(y, vec.back());
        vec.pop_back();
        break;
      }
    }
    vec.push_back(e);
  }
  return;
}
int main() {
  bool flag = 0;
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    if (c[i] == 2) flag = 1;
  }
  if (flag) nof();
  sort(c + 1, c + n + 1);
  reverse(c + 1, c + n + 1);
  if (c[1] != n) nof();
  if (n == 2) nof();
  if (n == 1) yesf();
  vec.push_back(pii(n - 1, 2));
  pos = -1;
  for (int i = n; i >= 1; i--)
    if (c[i] == 1) pos = i;
  leaf = n - pos + 1;
  tot_sum = n - 1;
  dfs(2);
  nof();
  return 0;
}

### Prompt

Your task is to create a CPP solution 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>
using namespace std;
using ll = long long;
int c[30];
int n, leaf, tot_sum, pos;
using pii = pair<int, int>;
void nof() {
  puts("NO");
  exit(0);
}
void yesf() {
  puts("YES");
  exit(0);
}
vector<pii> vec;
const double TL = 0.85;
void dfs(int k) {
  if (k == pos) {
    yesf();
    return;
  }
  vector<pii> V = vec;
  sort(V.begin(), V.end());
  if (clock() * 1.0 / CLOCKS_PER_SEC > TL) return;
  for (int i = 0; i < V.size(); i++) {
    if (i > 0 && V[i] == V[i - 1]) continue;
    if (clock() * 1.0 / CLOCKS_PER_SEC > TL) break;
    pii x = V[i];
    if (x.first < c[k]) continue;
    pii e, A = pii(-1, -1), B = pii(-1, -1);
    if (x.first == c[k] && x.second == 2) continue;
    for (auto &y : vec) {
      if (y == x) {
        swap(vec.back(), y);
        vec.pop_back();
        break;
      }
    }
    e = x;
    x.first -= c[k];
    x.second = 1;
    if (x.first) {
      vec.push_back(x);
      A = x;
    }
    B = pii(c[k] - 1, 2);
    vec.push_back(B);
    dfs(k + 1);
    if (A.first > 0) {
      for (auto &y : vec) {
        if (y == A) {
          swap(y, vec.back());
          vec.pop_back();
          break;
        }
      }
    }
    for (auto &y : vec) {
      if (y == B) {
        swap(y, vec.back());
        vec.pop_back();
        break;
      }
    }
    vec.push_back(e);
  }
  return;
}
int main() {
  bool flag = 0;
  scanf("%d", &n);
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    if (c[i] == 2) flag = 1;
  }
  if (flag) nof();
  sort(c + 1, c + n + 1);
  reverse(c + 1, c + n + 1);
  if (c[1] != n) nof();
  if (n == 2) nof();
  if (n == 1) yesf();
  vec.push_back(pii(n - 1, 2));
  pos = -1;
  for (int i = n; i >= 1; i--)
    if (c[i] == 1) pos = i;
  leaf = n - pos + 1;
  tot_sum = n - 1;
  dfs(2);
  nof();
  return 0;
}

```