#include <iostream>
using namespace std;
int main(){
	int n,k;
	while(cin>>n>>k,n){
		int A[100001]={0},tmp,flag=0,ans=0;
		for(int i=0;i<k;i++){
			cin>>tmp;
			if(tmp==0)flag=1;
			else A[tmp]=1;
		}
		for(int i=1;i<=n;i++)A[i]+=A[i-1];
		for(int i=1;i<=n;i++){
			for(int j=i+ans;j<=n;j++){
				if(A[j]-A[i-1]+flag<j-i+1)break;
				if((j-i+1)>ans)ans=(j-i+1);
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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

### Response

```cpp
#include <iostream>
using namespace std;
int main(){
	int n,k;
	while(cin>>n>>k,n){
		int A[100001]={0},tmp,flag=0,ans=0;
		for(int i=0;i<k;i++){
			cin>>tmp;
			if(tmp==0)flag=1;
			else A[tmp]=1;
		}
		for(int i=1;i<=n;i++)A[i]+=A[i-1];
		for(int i=1;i<=n;i++){
			for(int j=i+ans;j<=n;j++){
				if(A[j]-A[i-1]+flag<j-i+1)break;
				if((j-i+1)>ans)ans=(j-i+1);
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}

```