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>
#define MAX_AB 100000

int main() {
	unsigned long int yen[MAX_AB], max_drink[MAX_AB];
	int n, m;
	int money_sum=0;
	int i,j,temp;

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

	for (i = 0; i < n; i++) 
		scanf("%d %d", &yen[i], &max_drink[i]);

	for (i = 0; i < n - 1; i++) {
		for (j = n - 1; j > i; j--) {
			if (yen[j - 1] > yen[j]) {  /* 前の要素の方が大きかったら */
				temp = yen[j];        /* 交換する */
				yen[j] = yen[j - 1];
				yen[j - 1] = temp;

				temp = max_drink[j];        /* 交換する */
				max_drink[j] = max_drink[j - 1];
				max_drink[j - 1] = temp;
			}
		}
	}
		
	for (i = 0; m > 0;i++){
		if (max_drink[i] > m) {
			money_sum += yen[i] * max_drink[i];
			m -= max_drink[i];
		}
		else if (max_drink[i] < m && m>0) {
			money_sum += yen[i] * m;
			m = 0;
		}
	}
	printf("%d\n", money_sum);
	return 0;
}