#include <bits/stdc++.h>
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 100000005
#define MAX 100004
using namespace std;

ll a[MAX];
int dp[MAX][2];
int main(){
	int n, k, tmp;
	while(cin >> n >> k, n){
	fill(a, a+MAX, 0);
	fill((int*)dp, (int*)(dp+MAX), 0);
	for(int i = 0;i < k;i++){
		cin >> tmp;
		a[tmp] = 1;
	}
	
	int ans = 0;
	for(int i = 1;i < MAX;i++){
		if(a[i])dp[i][0] = dp[i-1][0]+1;
		if(dp[i-1][0] <= dp[i-1][1] && !a[i])dp[i][1] = dp[i-1][0]+1;
		else if(a[i])dp[i][1] = dp[i-1][1]+1;
		ans = max(ans, dp[i][0]);
		if(a[0])ans = max(ans, dp[i][1]);
	}
	//for(int i = 1;i <= 10;i++)cout << i << " " << dp[i][1]  << endl;
	cout << ans << endl;
	}
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <bits/stdc++.h>
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 100000005
#define MAX 100004
using namespace std;

ll a[MAX];
int dp[MAX][2];
int main(){
	int n, k, tmp;
	while(cin >> n >> k, n){
	fill(a, a+MAX, 0);
	fill((int*)dp, (int*)(dp+MAX), 0);
	for(int i = 0;i < k;i++){
		cin >> tmp;
		a[tmp] = 1;
	}
	
	int ans = 0;
	for(int i = 1;i < MAX;i++){
		if(a[i])dp[i][0] = dp[i-1][0]+1;
		if(dp[i-1][0] <= dp[i-1][1] && !a[i])dp[i][1] = dp[i-1][0]+1;
		else if(a[i])dp[i][1] = dp[i-1][1]+1;
		ans = max(ans, dp[i][0]);
		if(a[0])ans = max(ans, dp[i][1]);
	}
	//for(int i = 1;i <= 10;i++)cout << i << " " << dp[i][1]  << endl;
	cout << ans << endl;
	}
	return 0;
}
```