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 <bits/stdc++.h>
#include <unistd.h>
#define ll long long
#define ull unsigned long long
#define inf 1000000007
#define inf16 0x3f3f3f3f
#define INF 1000000000000000007LL
#define VI vector<int>
#define VPII vector<pair<int, int> >
#define VLL vector<ll>
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define st first
#define nd second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define endl '\n'
#define ALL(c) (c).begin(), (c).end()
#define flush fflush(stdout)
using namespace std;

#define N 100007

int n;

int a[N];

int32_t main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(NULL);

	cin >> n;

	for(int i = 1; i <= n; ++i)
	{
		cin >> a[i];
	}

	sort(a+1, a+1+n);

	#define cout cerr

	if(a[1]+1<a[n])
		cout << "No";
	else if(a[1]==a[n])
	{
		if(n>=2*a[1] || a[1]==n-1)
			cout << "Yes";
		else
			cout << "No";
	}
	else
	{
		int i = 1;

		while(a[i+1]==a[i])
			++i;

		if(n-i>=2*(a[i]-i+1))
			cout << "Yes";
		else
			cout << "No";
	}
}