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: Time Limit Exceeded)

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

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    int *a, *b;
    a = (int *)calloc(n, sizeof(int));
    b = (int *)calloc(n, sizeof(int));
    for (int i=0; i<n; i++) {
	scanf("%d %d", &a[i], &b[i]);
    }
    int drink = 0;
    long sum = 0;
    while (drink < m) {
	int min = INT_MAX;
	int minin;
	for (int i=0; i<n; i++) {
	    if (min > a[i] && b[i] != 0) {
		minin = i;
		min = a[i];
	    }
	}
	while (b[minin] > 0 && drink < m) {
	    b[minin]--;
	    drink++;
	    sum += a[minin];
	}
    }
    printf("%ld\n", sum);
}