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: Runtime Error)

#include<stdio.h>

int main(void)
{
        int n, m, i, j, a[10000], b[10000], tmp, k=0;
        double r = 0;

        scanf("%d%d", &n,&m);

        for(i = 0; i < n; i++) {
                scanf("%d%d", &a[i], &b[i]);
        }

        for (i = 0; i < n; i++) {
                for (j=i+1; j<n; ++j) {
                        if (a[i] >= a[j]) {
                                tmp =  a[i];
                                a[i] = a[j];
                                a[j] = tmp;
                                tmp = b[i];
                                b[i] = b[j];
                                b[j] = tmp;
                        }
                }
        }

        for(i = 0; i < n; i++) {
                for(j = 0; j < b[i]; j++) {
                        r += a[i];
                        k++;
                        if(k == m) {
                                printf("%.0lf", r);
                                return 0;
                        }
                }
        }
}