#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
	while (1) {
		int k, n, num[100010] = {}, pre = 1, now_max = 0, flag = 0, numorg;
		cin >> n >> k;
		if (n == 0 && k == 0) {
			break;
		}
		for (int i = 0; i < k; i++) {
			int a;
			cin >> a;
			num[a] = 1;

		}
		numorg = num[0];
		for (int i = 1; i <= n; i++) {
			if (num[i] == 1) {
				if (num[i + 1] == 1) {
					pre++;
				}
				else {
					if (num[i + 2] == 1 && num[0] == 1) {
						pre += 2;
						num[0] = 0;
					}
					else {
						now_max = max(pre, now_max);
						pre = 1;
						num[0] = numorg;
					}
				}
			}
		}

		cout << now_max << endl;
	}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<algorithm>

using namespace std;

int main()
{
	while (1) {
		int k, n, num[100010] = {}, pre = 1, now_max = 0, flag = 0, numorg;
		cin >> n >> k;
		if (n == 0 && k == 0) {
			break;
		}
		for (int i = 0; i < k; i++) {
			int a;
			cin >> a;
			num[a] = 1;

		}
		numorg = num[0];
		for (int i = 1; i <= n; i++) {
			if (num[i] == 1) {
				if (num[i + 1] == 1) {
					pre++;
				}
				else {
					if (num[i + 2] == 1 && num[0] == 1) {
						pre += 2;
						num[0] = 0;
					}
					else {
						now_max = max(pre, now_max);
						pre = 1;
						num[0] = numorg;
					}
				}
			}
		}

		cout << now_max << endl;
	}
	return 0;
}
```