#include<iostream>
#include<algorithm>
#define F 100000
using namespace std;
int c[F],s[F],e[F];
int main(){
	int n,k,qtop;
	while(true){
	cin>>n>>k;
	if(!n&&!k)break;
	for(int i=0;i<k;i++)cin>>c[i];
	sort(c,c+k);
	bool t=false;
	if(!c[0])t=true;
	qtop=1,s[0]=c[t];
	for(int i=1+t;i<k;i++){
		if(c[i]!=c[i-1]+1){
			e[qtop-1]=c[i-1];
			s[qtop]=c[i];
			qtop++;
		}
	}
	e[qtop-1]=c[k-1];
	int ans=0;
	for(int i=0;i<qtop;i++)ans=max(ans,e[i]-s[i]+1);
	if(t){
		ans=min(ans+1,k);
		for(int i=0;i<qtop-1;i++)if(e[i]+2==s[i+1])ans=max(ans,e[i]+e[i+1]-s[i]-s[i+1]+3);
	}
	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>
#include<algorithm>
#define F 100000
using namespace std;
int c[F],s[F],e[F];
int main(){
	int n,k,qtop;
	while(true){
	cin>>n>>k;
	if(!n&&!k)break;
	for(int i=0;i<k;i++)cin>>c[i];
	sort(c,c+k);
	bool t=false;
	if(!c[0])t=true;
	qtop=1,s[0]=c[t];
	for(int i=1+t;i<k;i++){
		if(c[i]!=c[i-1]+1){
			e[qtop-1]=c[i-1];
			s[qtop]=c[i];
			qtop++;
		}
	}
	e[qtop-1]=c[k-1];
	int ans=0;
	for(int i=0;i<qtop;i++)ans=max(ans,e[i]-s[i]+1);
	if(t){
		ans=min(ans+1,k);
		for(int i=0;i<qtop-1;i++)if(e[i]+2==s[i+1])ans=max(ans,e[i]+e[i+1]-s[i]-s[i+1]+3);
	}
	cout<<ans<<endl;
	}
	return 0;
}
```