Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

typedef struct{
    long price;
    int num;
}shop;

//昇順
int asc(const void* a, const void* b){
    shop p = *(shop*)a;
    shop q = *(shop*)b;

    if(p.price<q.price) return -1;
    else if(p.price>q.price) return 1;
    return 0;
}

int main(void){
    int N,M;
    scanf("%d%d",&N,&M);

    //S[0]~S[N-1]に格納する
    shop* S;
    NEW(S,N);
    for(int i=0;i<N;i++){
        scanf("%ld%d",&S[i].price, &S[i].num);
    }
    qsort(S,N,sizeof(shop),asc);

    long ans=0;
    int i=0;
    while(M>0){
        if(M>=S[i].num){
            ans+=S[i].price*S[i].num;
            M-=S[i].num;
        }
        else{
            ans+=S[i].price*M;
            M=0;
        }
    }
    printf("%ld\n",ans);

    return 0;
}