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 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#define LOCAL

#include <fstream>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <vector>
#include <climits>
#include <queue>
#include <assert.h>
#include <set>
#include <numeric>
#include <memory.h>

#define FOR(i,a,b) for(int i=a; i<b; i++)
#define REP(i,n) FOR(i,0,n)

//#define int long long
typedef long long ll;

const ll INF = 1e18;

using namespace std;

int N;
int a[100010];

signed main()
{
#ifdef LOCAL
    ifstream in("input.txt");
    cin.rdbuf(in.rdbuf());
#endif

    cin >> N;
    int mx = 0, mn = 999999;
    REP(i,N) {
        cin >> a[i];
        mx = max(mx, a[i]);
        mn = min(mn, a[i]);
    }

    if (mx > mn + 1) {
        cout << "No" << endl;
        return 0;
    }

    if (mx == mn) {
        if (mx == N - 1 || mx <= N / 2) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
    } else {
        int p = 0, q = 0;
        REP(i,N) {
            if (a[i] == mx) p++;
            else q++;
        }
        if (p == 1) {
            cout << "No" << endl;
            return 0;
        }
        if (p / 2 + q >= mx) {
            cout << "Yes" << endl;
        } else {
            cout << "No" << endl;
        }
    }
}