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: Accepted)

#include <ctime>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <map>
using namespace std;

typedef long long ll;
typedef double lf;
typedef long double Lf;
typedef pair <int,int> pii;
typedef pair <ll, ll> pll;

#define TRACE(x) cerr << #x << "  " << x << endl
#define FOR(i, a, b) for (int i = (a); i < int(b); i++)
#define REP(i, n) FOR(i, 0, n)
#define all(x) (x).begin(), (x).end()
#define _ << " " <<

#define fi first
#define sec second
#define mp make_pair

const int MAXN = 100100;

int n, p[MAXN], cnt[MAXN];

int main() {
	scanf("%d",&n);
	REP(i, n) scanf("%d",&p[i]);
	REP(i, n) cnt[p[i]]++;

  int tot = 0;
  REP(i, MAXN) tot += (cnt[i] > 0);

  if (tot == 1) {
    (2 * p[0] > n && p[0] != n - 1) ? puts("No") : puts("Yes");
    return 0;
  }

  if (tot > 2) {
    puts("No");
    return 0;
  }

  sort(p, p + n);
  int a = p[0];
  int b = p[n - 1];

  if (b - a != 1 || !(cnt[a] + 1 <= b && b <= cnt[a] + cnt[b]/2)) puts("No");
  else puts("Yes");
	return 0;
}