CUIT ACM Personal Training 11.27(FM）D - Yaroslav and Permutations

D - Yaroslav and Permutations

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Description

Yaroslav has an array that consists of n integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.

Help Yaroslav.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array. The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000) — the array elements.

Output

In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.

Sample Input

Input

1

1

Output

YES

Input

3

1 1 2

Output

YES

Input

4

7 7 7 7

Output

NO

Hint

In the first sample the initial array fits well.

In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.

In the third sample Yarosav can't get the array he needs.

題解：此題的題意是給你一串數字，讓你判斷能否通過交換，使得他們相鄰彼此不相同。

那麼這個其實就是一個模擬插空的過程，首先挑出出現次數最多的數字，然後將其排成一排，用其他數字一個一個去插空，那麼這些數字就有可能符合題意。

情況一：6 5 4 5 3   5出現的次數最多，數量2<=(5+1)/2,所以成立；

情況二：5 4 5 4 5   5出現的次數最多，數量3<=(5+1)/2,所以成立；

情況二：5 4 5 5 5   5出現的次數最多，數量4>(5+1)/2,所以不成立；

所以我們可以很清楚地看出來(n+1)/2是判斷點。

AC代碼如下:

#include <iostream>
#include <cstdio>
using namespace std;

int n,A[1001],temp;
int maxnum;

int main()
{
	scanf("%d",&n);
	maxnum=0;
	for(int i=0;i<n;i++){
		scanf("%d",&temp); A[temp]++;
		if(A[temp]>maxnum) maxnum=A[temp];
	}
	if(maxnum>(n+1)/2) printf("NO\n");
	else printf("YES\n");
	return 0;
}