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)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <functional>
#include <numeric>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <utility>
#include <sstream>
#include <complex>
#include <fstream>
#include <bitset>
#include <time.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef vector<ll> V;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = (1LL << 31) - 1;
const ll MOD = 1e9 + 7;

#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,N) for(int i=0;i<(N);i++)
#define ALL(s) (s).begin(),(s).end()
#define EQ(a,b) (abs((a)-(b))<EPS)
#define EQV(a,b) ( EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()) )
#define fi first
#define se second
#define N_SIZE (1LL << 20)
#define NIL -1
#define MAX_N 100100 * 3

ll n;
ll a[101000];

int main() {
	cin >> n;
	REP(i, n)cin >> a[i];
	REP(i, n) {
		if (a[0] == a[i] && a[i] == 1) {
			if (i == n - 1) {
				cout << "Yes" << endl;
				return 0;
			}
		}
		else break;
	}
	REP(i, n) {
		if (a[0] == a[i] && a[i] == n - 1) {
			if (i == n - 1) {
				cout << "Yes" << endl;
				return 0;
			}
		}
		else break;
	}
	ll sum = 0;
	REP(i, n)sum += a[i];
	if (sum >= n&&sum <= (n*(n - 1) - (n - 2))) {
		if (((sum - (2 * n - 1)) % (n - 1) == 0) || ((sum - (2 * n - 1)) % (n - 1) == 1)) {
			cout << "Yes" << endl;
		}
		else cout << "No" << endl;
	}
	else cout << "No" << endl;
}