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>

using namespace std;

int main() {
    while(true) {
        int n, k;
        cin >> n >> k;
        if(!n && !k) break;
        vector<bool> v(n+1, false);

        bool wildcard = false;
        for(int i = 0; i < k; ++i) {
            int num;
            cin >> num;
            v[num] = true;
            if(num == 0) wildcard = true;
        }

        int prevlen = 0;
        int currlen = 0;
        int ans = 0;
        int i = 1;
        while(i <= n) {
            while(!v[i] && i <= n) ++i;
            if(i >= n+1) break;
            while(v[i++]) ++currlen;
            if(wildcard) {
                ans = max(ans, prevlen+currlen+1);
                if(i+1 <= n && v[i+1]) prevlen = currlen;
                else prevlen = 0;
            }
            else {
                ans = max(ans, currlen);
            }
            currlen = 0;
        }

        cout << ans << endl;
    }
    return 0;
}