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
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

void solve()
{
	int n, k;
	while (cin >> n >> k, n || k)
	{
		bool isBlank = false;
		vector<int> card(k);
		for (int i = 0; i < k; ++i)
		{
			cin >> card[i];
			if (card[i] == 0)
			{
				isBlank = true;
			}
		}
		sort(card.begin(), card.end());
		int pos = 1;
		if (card[0] == 0)
		{
			++pos;
		}
		bool isUsed;
		int ans = 0;
		for (int i = pos; i < k; ++i)
		{
			if (isBlank)
			{
				isUsed = false;
			}
			else
			{
				isUsed = true;
			}
			int count = 1;
			int pre = card[i - 1];
			for (int j = i; j < k; ++j)
			{
				if (pre == card[j] - 1)
				{
					++count;
					ans = max(ans, count);
					pre = card[j];
				}
				else
				{
					if (isUsed)
					{
						count = 0;
						break;
					}
					else
					{
						isUsed = true;
						++count;
						ans = max(ans, count);
						pre = card[j] - 1;
						--j;
					}
				}
			}
		}
		cout << ans << endl;
	}
}

int main()
{
	solve();
	return(0);
}