Problem:
Score : 700 points

Problem Statement

There are N cats.
We number them from 1 through N.

Each of the cats wears a hat.
Cat i says: "there are exactly a_i different colors among the N - 1 hats worn by the cats except me."

Determine whether there exists a sequence of colors of the hats that is consistent with the remarks of the cats.

Constraints

2 ≤ N ≤ 10^5

1 ≤ a_i ≤ N-1

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print Yes if there exists a sequence of colors of the hats that is consistent with the remarks of the cats; print No otherwise.

Sample Input 1

3
1 2 2

Sample Output 1

Yes

For example, if cat 1, 2 and 3 wears red, blue and blue hats, respectively, it is consistent with the remarks of the cats.

Sample Input 2

3
1 1 2

Sample Output 2

No

From the remark of cat 1, we can see that cat 2 and 3 wear hats of the same color.
Also, from the remark of cat 2, we can see that cat 1 and 3 wear hats of the same color.
Therefore, cat 1 and 2 wear hats of the same color, which contradicts the remark of cat 3.

Sample Input 3

5
4 3 4 3 4

Sample Output 3

No

Sample Input 4

3
2 2 2

Sample Output 4

Yes

Sample Input 5

4
2 2 2 2

Sample Output 5

Yes

Sample Input 6

5
3 3 3 3 3

Sample Output 6

No

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <algorithm>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define repi(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,a) repi(i,0,a)
#define all(a) (a).begin(), (a).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

using ll = long long;

ll N;
ll a[100010];
ll mi = 1ll<<60, ma = 0;

int main()
{
  std::cin >> N;

  rep( i, N )
  {
    std::cin >> a[i];
    chmin( mi, a[i] );
    chmax( ma, a[i] );
  }

  ll cnt = 0;

  rep( i, N )
    cnt += a[i] == ma;

  if( ma == mi && (mi == N-1 || 2*mi <= N) )
  {
    std::cout << "Yes" << std::endl;

    return 0;
  }

  ll x = N-cnt, y = cnt;
  ll A = ma;

  std::cout << (x < A && 2*(A-x) <= y ? "Yes" : "No") << std::endl;

  return 0;
}