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: Accepted)

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

int comp(const void* a, const void* b){
    long int A = *(long int*)a;
    long int B = *(long int*)b;

    if(A > B){return 1;}
    if(A == B){return 0;}
    return -1;
}


int main(void){
    int N, M;
    scanf("%d %d", &N, &M);
    long int a[N][2];
    int i,j;
    for(i = 0; i < N; i++){scanf("%ld %ld", &a[i][0], &a[i][1]);}
    //for(i = 0; i < N; i++){for(j = 0; j < 2; j++){printf("%ld  ", a[i][j]);}}
    //return 0;
    qsort(a, N, sizeof(a[0]), comp);
    //for(i = 0; i < N; i++){printf("値段：%ld 本数:%ld\n", a[i][0], a[i][1]);}

    long int r = M;//買うべき残りの本数
    long int ans = 0;
    i = 0;
    while(r > 0){
        if(a[i][1] > r){
            ans += a[i][0] * r;
            break;
        }else{
            ans += a[i][0] * a[i][1];
            r -= a[i][1];
        }
        i++;
    }
    printf("%ld\n", ans);
    return 0;
}