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>
int main(void){
    int N, M = 0;
    scanf("%d %d", &N, &M);
    int A[100000] = {0}; int B[10000] = {0};
    for(int i = 0; i < N; i++){
        scanf("%d %d", &A[i], &B[i]);
    }

    int total = 0; long sum = 0;
    while(M - total > 0){
        int min = 100000; int minno = 0;
        for(int i = 0; i < N; i++){
            if(min > A[i]){
                min = A[i];
                minno = i;
            }
        }
        // 最安値の店を見つける
        if(M - total > 0){
            if(M - total > B[minno]){
                sum += A[minno] * B[minno];
                total += B[minno];
            }else {
                sum += A[minno] * (M - total);
                total += M - total;
            }
        }
        A[minno] = 100000000; B[minno] = 1000000;
    }

    printf("%d\n", sum);
    return 0;
}