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

	for (i = 0; i<N-1; i++) {
		for (j = N-1; j > i; j--) {
			if (A[j] < A[j -1]) {
				int a = A[j], b = B[j];
				A[j] = A[j -1];
				A[j -1] = a;
				B[j] = B[j -1];
				B[j - 1] = b;
			}
		}
	}

	int sum = 0, money = 0;
	for (i = 0; i < N; i++) {
		for (j = 0; j < B[i]; j++) {
			sum ++;
			money += A[i];
			if (sum == M)break;
		}
		if (sum == M)break;
	}
	printf("%d\n", money);
	return 0;
}