（CodeForces

（CodeForces - 609A）USB Flash Drives

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

Sean is trying to save a large file to a USB flash drive. He has n USB flash drives with capacities equal to a1, a2, …, an megabytes. The file size is equal to m megabytes.

Find the minimum number of USB flash drives needed to write Sean’s file, if he can split the file between drives.

Input

The first line contains positive integer n (1 ≤ n ≤ 100) — the number of USB flash drives.

The second line contains positive integer m (1 ≤ m ≤ 105) — the size of Sean’s file.

Each of the next n lines contains positive integer ai (1 ≤ ai ≤ 1000) — the sizes of USB flash drives in megabytes.

It is guaranteed that the answer exists, i. e. the sum of all ai is not less than m.

Output

Print the minimum number of USB flash drives to write Sean’s file, if he can split the file between drives.

Examples

Input

3
5
2
1
3

Output

2

Input

3
6
2
3
2

Output

3

Input

2
5
5
10

Output

1

Note

In the first example Sean needs only two USB flash drives — the first and the third.

In the second example Sean needs all three USB flash drives.

In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second.

题目大意：有n个USB，每个USB可以存放a[i]字节数的文件，现在需要存放m字节的文件，问最少需要最少个USB。

思路：肯定是容量大的USB先装，然后在装小的，一直到装够m为止。

#include<cstdio>#include<algorithm>using namespace std;const int maxn=105;int a[maxn];bool cmp(int a,int b){    return a>b;}int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        for(int i=0;i<n;i++) scanf("%d",a+i);        sort(a,a+n,cmp);        int ans,sum=0;        for(int i=0;i<n;i++)        {            sum+=a[i];            if(sum>=m)            {                ans=i+1;                break;            }        }        printf("%d\n",ans);    }    return 0;}