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 <stdlib.h>
#include <string.h>
#include <math.h>

/*
void sort(int *a, int *b)
{
	int tmp;
	int i, j, k;
	for (i = 0; i < sizeof(a) / sizeof(int); i++) {
		for (j = i + 1; j < sizeof(a) / sizeof(int); j++) {
			if (a[i] > a[j]) {
				int *small_a, *large_a, *small_b, *large_b;
				int tmp_a, tmp_b;
				small_a = &a[i];
				large_a = &a[j];
				tmp_a = a[i];
				
				small_b = &b[i];
				large_b = &b[j];
				tmp_b = b[i];

				*small_a = *large_a;
				*large_a = tmp_a;
				
				*small_b = *large_b;
				*large_b = tmp_b;
			}
		}
	}

}
この関数には配列のコピーが渡されているので、大小比較の部分において
全ての場合で入力された順番どおりの比較が行われている
*/

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

//	sort(A, B);
	
	for (i = 0; i < N; i++) {
		for (j = i + 1; j < N; j++) {
			if (A[i] > A[j]) {
				int tmp_A, tmp_B;
				tmp_A = A[i];
				tmp_B = B[i];

				A[i] = A[j];
				A[j] = tmp_A;
				B[i] = B[j];
				B[j] = tmp_B;
			}
		}
	}
	
//	printf("sizeof(a) / sizeof(int): %ld\n", sizeof(A) / sizeof(int));
//	for (i = 0; i < N; i++) 
//		printf("A[%d]: %d, B[%d]: %d\n", i, A[i], i, B[i]);

	long long int cost;
	int num;
	for (cost = 0, num = 0, i = 0; num < M; num++) {
		if (B[i] == 0) {
			i++;
		} 
		cost += A[i];
		B[i]--;
		
//		printf("cost: %lld, A[%d]: %d, B[%d]: %d\n", cost, i, A[i], i, B[i]);

	}

	printf("%lld\n", cost);

	return 0;
}