codeforces 1015C

C. Songs Compression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Ivan has $\mathrm{nn songs\; on\; his\; phone.\; The\; size\; of\; the ii-th\; song\; is aiai bytes.\; Ivan\; also\; has\; a\; flash\; drive\; which\; can\; hold\; at\; most mm bytes\; in\; total.\; Initially,\; his\; flash\; drive\; is\; empty.}$

Ivan wants to copy all $\mathrm{nn songs\; to\; the\; flash\; drive.\; He\; can\; compress\; the\; songs.\; If\; he\; compresses\; the ii-th\; song,\; the\; size\; of\; the ii-th\; song\; reduces\; from aiai to bibi bytes\; (bi<aibi<ai).}$

Ivan can compress any subset of the songs (possibly empty) and copy all the songs to his flash drive if the sum of their sizes is at most $\mathrm{mm.\; He\; can\; compress any subset\; of\; the\; songs\; (not\; necessarily\; contiguous).}$

Ivan wants to find the minimum number of songs he needs to compress in such a way that all his songs fit on the drive (i.e. the sum of their sizes is less than or equal to $\mathrm{mm).}$

If it is impossible to copy all the songs (even if Ivan compresses all the songs), print "-1". Otherwise print the minimum number of songs Ivan needs to compress.

Input

The first line of the input contains two integers $\mathrm{nn and mm (1\le n\le 105,1\le m\le 1091\le n\le 105,1\le m\le 109)\; —\; the\; number\; of\; the\; songs\; on\; Ivan\text{'}s\; phone\; and\; the\; capacity\; of\; Ivan\text{'}s\; flash\; drive.}$

The next $\mathrm{nn lines\; contain\; two\; integers\; each:\; the ii-th\; line\; contains\; two\; integers aiai and bibi (1\le ai,bi\le 1091\le ai,bi\le 109, ai>biai>bi)\; —\; the\; initial\; size\; of\; the ii-th\; song\; and\; the\; size\; of\; the ii-th\; song\; after\; compression.}$

Output

If it is impossible to compress a subset of the songs in such a way that all songs fit on the flash drive, print "-1". Otherwise print the minimum number of the songs to compress.

Examples

input

Copy

4 21
10 8
7 4
3 1
5 4

output

Copy

2

input

Copy

4 16
10 8
7 4
3 1
5 4

output

Copy

-1

题意：给你n件物品，和一个空间为m的背包，这n件物品的初始占用空间为ai，其占用空间可以缩小为bi，但是需要花费ai-bi的花费，求在花费最小的情况下有多少物品不用被压缩
题解：我们换个方向，这n个物品被压缩后的大小为bi，先全部加起来，如果压缩后的物品空间大小小于空间m，那么就可以把所有的物品加入背包，为了使花费最小，
　　　我们剩下的空间必须尽量装满，所以我们就将花费按照从小到大排序，一个个装，装到不能装了后就跳出来就行

代码如下：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int maxn = 1e5+5;
LL a[maxn];
LL b[maxn];
LL c[maxn];
int main(){
#ifndef ONLINE_JUDGE
    FIN
#endif
    LL n,m;
    LL sumb=0;
    int flag=1;
    scanf("%lld%lld",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%lld%lld",&a[i],&b[i]);
        c[i]=a[i]-b[i];
        sumb+=b[i];
    }
    if(sumb>m){
        cout<<"-1"<<endl;
    }else{
        sort(c, c + n);
        int ans = 0;
        for (int i = 0; i < n; i++) {
            if (sumb + c[i] <= m) {
                sumb += c[i];
                ans++;
            }
            else break;
        }
        printf("%lld\n", n - ans);
    }

}