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: Accepted)

#include <stdio.h>

void  SWAP(int *x, int *y) {
  int temp;
  temp = *x;
  *x = *y;
  *y = temp;
}

void quicksort(int *array, int *array2, int left, int right) {
    int l = left;
    int r = right;
    int pivot = array[(left + right) / 2];

    while (1) {
        while (array[l] < pivot) l++;
        while (array[r] > pivot) r--;
        if (l > r) break;
        SWAP(&array[l], &array[r]);
        SWAP(&array2[l], &array2[r]);
        l++, r--;
    };

    if (left < r)  quicksort(array, array2, left, r);
    if (l < right) quicksort(array, array2, l, right);
}

int main(void){
    int N, M;
    scanf("%d %d", &N, &M);
    int A[N];
    int B[N];

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

    quicksort(A, B, 0, N-1);

    long sum = 0;
    int num = 0;
    int tmp = 0;
    int i = 0;

    while(num < M){
        num += B[i];
        sum += (long)A[i] * B[i];
        tmp = i;
        i++;
    }
    sum -= (long)(num - M) * A[tmp];
    
    printf("%ld\n", sum);
    return 0; 
}