#include <iostream>
#include <vector>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
int solve(vector<int> data,int white,int n){
	int dp[100001] = {0};
	rep(i,100001)dp[i] = 1;
	rep(i,data.size()) dp[data[i]] = 0;
	
	dp[0] = 0;
	for(int i=1;i<=n;i++)
		dp[i] = dp[i-1] + dp[i];
	
	int l = 1 , r = 1 , ans = 0;
	while(r <= n){
		if(dp[r]-dp[l-1] <= white){
			ans = max(ans,r-l+1);
			r++;
		}else{
			l++;
		}
	}
	return ans;
}
int main(){
	int n,k;
	while(cin >> n >> k , n){
		vector<int> data;
		int e = 0;
		rep(i,k){
			int t; cin >> t;
			if(t)data.push_back(t);
			else e++;
		}
		cout << solve(data,e,n) << endl;
	}
	
}
### Prompt

Please formulate 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 <vector>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
int solve(vector<int> data,int white,int n){
	int dp[100001] = {0};
	rep(i,100001)dp[i] = 1;
	rep(i,data.size()) dp[data[i]] = 0;
	
	dp[0] = 0;
	for(int i=1;i<=n;i++)
		dp[i] = dp[i-1] + dp[i];
	
	int l = 1 , r = 1 , ans = 0;
	while(r <= n){
		if(dp[r]-dp[l-1] <= white){
			ans = max(ans,r-l+1);
			r++;
		}else{
			l++;
		}
	}
	return ans;
}
int main(){
	int n,k;
	while(cin >> n >> k , n){
		vector<int> data;
		int e = 0;
		rep(i,k){
			int t; cin >> t;
			if(t)data.push_back(t);
			else e++;
		}
		cout << solve(data,e,n) << endl;
	}
	
}
```