#include<cstdio>
#include<cstring>
#define max(a,b) (a>b?a:b)
int main(){
	int k[100001],a,b,table[2][100001];
	while(scanf("%d%d",&a,&b),a,b){
	memset(k,0,sizeof(k));
	memset(table,0,sizeof(table));
		for(int i=0;i<b;i++){
			int c;
			scanf("%d",&c);k[c]=1;
		}	
		int res=0;
		for(int i=1;i<=a;i++){
			if(k[i]){
				table[0][i]=table[0][i-1]+1;
				table[1][i]=table[1][i-1]+1;
			}
			else{
				if(k[0]){
					table[1][i]=table[0][i-1]+1;
				}
			}
			res=max(res,max(table[0][i],table[1][i]));
		}
		printf("%d\n",res);
	}
	return 0;
}
### Prompt

Your challenge is to write 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<cstdio>
#include<cstring>
#define max(a,b) (a>b?a:b)
int main(){
	int k[100001],a,b,table[2][100001];
	while(scanf("%d%d",&a,&b),a,b){
	memset(k,0,sizeof(k));
	memset(table,0,sizeof(table));
		for(int i=0;i<b;i++){
			int c;
			scanf("%d",&c);k[c]=1;
		}	
		int res=0;
		for(int i=1;i<=a;i++){
			if(k[i]){
				table[0][i]=table[0][i-1]+1;
				table[1][i]=table[1][i-1]+1;
			}
			else{
				if(k[0]){
					table[1][i]=table[0][i-1]+1;
				}
			}
			res=max(res,max(table[0][i],table[1][i]));
		}
		printf("%d\n",res);
	}
	return 0;
}
```