Codeforce 915A Garden

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Luba thinks about watering her garden. The garden can be represented as a segment of length k. Luba has got n buckets, the i-th bucket allows her to water some continuous subsegment of garden of length exactly a**i each hour. Luba can’t water any parts of the garden that were already watered, also she can’t water the ground outside the garden.

Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length a**i if she chooses the i-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.

See the examples for better understanding.

Input

The first line of input contains two integer numbers n and k (1 ≤ n, k ≤ 100) — the number of buckets and the length of the garden, respectively.

The second line of input contains n integer numbers a**i (1 ≤ a**i ≤ 100) — the length of the segment that can be watered by the i-th bucket in one hour.

It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.

Output

Print one integer number — the minimum number of hours required to water the garden.

Examples

input

1
2
3 6
2 3 5

output

1
2

input

1
2
6 7
1 2 3 4 5 6

output

1
7

Note

In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can’t choose the bucket that allows to water the segment of length 5 because then we can’t water the whole garden.

In the second test we can choose only the bucket that allows us to water the segment of length 1.

C++ Code

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#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>

using namespace std;

int main()
{
    int n, k;
    int tmp;
    int mi;
    while(~scanf("%d%d", &n, &k))
    {
        mi = 1000;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &tmp);
            if(k % tmp == 0)
            {
                mi = min(k / tmp, mi);
            }
        }
        printf("%d\n", mi);
    }
    return 0;
}