problem

There are n cards with one integer from 1 to n and one blank card. Of these n + 1 cards, k cards are given, but 1 ≤ k ≤ n. You can write one integer from 1 to n on a blank card. I want to make a continuous sequence of integers as long as possible with just a given card.

Write a program that outputs the maximum length of a contiguous sequence of integers that can be made from a given card when the given card is entered.



input

The input consists of multiple datasets. Each dataset is given in the following format. The input ends on a line containing two zeros.

On the first line, two integers n (1 ≤ n ≤ 100000) and k (1 ≤ k ≤ n) are written in this order, separated by one blank. The following k line contains one integer. Written one by one, representing the integers written on the given k cards. Blank cards are represented by 0.

Of the scoring data, 40% of the points are 1 ≤ n ≤ 1000, 1 ≤ k ≤ 500, 20% of the points are 1 ≤ n ≤ 60000, 1 ≤ k ≤ 50000. Satisfy 1 ≤ n ≤ 100000, 1 ≤ k ≤ 100000.

The number of datasets does not exceed 5.

output

Outputs an integer on one line for each dataset.

Examples

Input

7 5
6
2
4
7
1
7 5
6
2
0
4
7
0 0


Output

2
4


Input

None


Output

None
#include <iostream>
#include <vector>
#include <cstring>
int main()
{
	int n,k;
	while(std::cin>>n>>k,n|k){
		int memo[100002];
		memset(memo,0,sizeof(memo));
		for(int i=0; i<k; i++){
			int a; std::cin>>a;
			memo[a]=1;
		}
		int cnt=0;
		int flag=0;
		int ans = 0;
		int tmp = -1;
		std::vector<int> v;
		for(int i=1; i<=n+1; i++){
			if      ( memo[i] ){
				memo[i]+=memo[i-1],cnt++;
				if( ans<memo[i] )ans=memo[i];
				flag=1;
			}else if( flag ){
				memo[ i-memo[i-1] ] = memo[i-1];
				flag=0;
			}
			if( memo[0] && !memo[i] && memo[i-1] && memo[i+1] ){
				v.push_back(i);
			}
			if( cnt==k && tmp==-1 )tmp=i;
			if( cnt==k && tmp<i )break;
		}
		for(int i=0; i<v.size() && memo[0]; i++){
			if( ans<memo[v[i]-1]+memo[v[i]+1]+1 )
				ans = memo[v[i]-1]+memo[v[i]+1]+1;
		}
		/*
		for(int i=0; i<=n; i++){
			printf("%d,",memo[i]);
		}
		puts("");
		*/
		printf("%d\n",ans);
	}
	return 0;
}