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){

	long long int N, M;
	scanf("%lld", &N);
	scanf("%lld", &M);

	long long int A[10000][2];
	long long int i, j;

	for(i = 0; i < N; i++){
		for(j = 0; j < 2; j++){
			scanf("%lld", &A[i][j]);
		}
	}
	
	long long int temp1 = 0, temp2 = 0;

	for(i = 0; i < N - 1; i++){
		if(A[i][0] < A[i + 1][0]){
			temp1 = A[i][0];
			A[i][0] = A[i + 1][0];
			A[i + 1][0] = temp1;

			temp2 = A[i][1];
                        A[i][1] = A[i + 1][1];
                        A[i + 1][1] = temp2;
			
		}
	}

	long long int ans = 0;
	long long int count = 0;

	for(i = 0; i < N; i++){
		count = count + A[i][1];

		if(count < M){
			ans = ans + A[i][0] * A[i][1];
		}else if(count == M){
			ans = ans + A[i][0] * A[i][1];
			break;
		}else{
			ans = ans + A[i][0] * (M - (count - A[i][1]));
			break;
		}
	}

	printf("%lld\n", ans);
	return 0;

}