Educational Codeforces Round 3 A

A. 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.

Sample test(s)

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.

贪心，这个没什么好说的，拿大的放进去就好

#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int i,j;
int t;
int n,m;
int a[1000000];
int main()
{
    int sum=0;
    int num=1;
    cin>>n>>m;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n);
    for(i=n-1;i>=0;i--)
    {
        sum+=a[i];
        if(sum>=m)
        {
            break;
        }
    }
    cout<<n-i<<endl;
    return 0;
}