#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;

vector<pair<int, int> > v;
int p[111111];

int main(void)
{
	while (1)
	{
		int n, k, i, a, white = 0;

		v.clear();
		for (i = 1; i <= 100000; i++)
			p[i] = 0;

		scanf("%d %d" ,&n, &k);

		if (n == 0 && k == 0)
			break;

		for (i = 0; i < k; i++)
		{
			scanf("%d" ,&a);

			!a ? white = 1 : p[a] = 1;
		}

		for (i = 1; i <= n; i++)
			if (p[i])
			{
				if (!p[i - 1])
					v.push_back(make_pair(i, i));
				else
					v.back().second = i;
			}

		int ans = 0;
		for (i = 0; i < v.size(); i++)
			ans = max(ans, v[i].second - v[i].first + 1);

		if (white)
		{
			++ans;
			for (i = 0; i < v.size() - 1; i++)
				if (v[i + 1].first - v[i].second == 2)
					ans = max(ans, v[i + 1].second - v[i].first + 1);
		}

		printf("%d\n" ,ans);
	}
	return 0;
}


### Prompt

Develop a solution in cpp to the problem described below:

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 <algorithm>
#include <vector>
using namespace std;

vector<pair<int, int> > v;
int p[111111];

int main(void)
{
	while (1)
	{
		int n, k, i, a, white = 0;

		v.clear();
		for (i = 1; i <= 100000; i++)
			p[i] = 0;

		scanf("%d %d" ,&n, &k);

		if (n == 0 && k == 0)
			break;

		for (i = 0; i < k; i++)
		{
			scanf("%d" ,&a);

			!a ? white = 1 : p[a] = 1;
		}

		for (i = 1; i <= n; i++)
			if (p[i])
			{
				if (!p[i - 1])
					v.push_back(make_pair(i, i));
				else
					v.back().second = i;
			}

		int ans = 0;
		for (i = 0; i < v.size(); i++)
			ans = max(ans, v[i].second - v[i].first + 1);

		if (white)
		{
			++ans;
			for (i = 0; i < v.size() - 1; i++)
				if (v[i + 1].first - v[i].second == 2)
					ans = max(ans, v[i + 1].second - v[i].first + 1);
		}

		printf("%d\n" ,ans);
	}
	return 0;
}


```