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>
#include <stdlib.h>

int kazu[100001] = {0};

void swap (long long *x, long long *y) {
  long long temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

void swap2 (int *x,  int *y) {
  int temp;    // 値を一時保存する変数

  temp = *x;
  *x = *y;
  *y = temp;
}

int partition (long long array[], int left, int right) {
  int i, j, pivot;
  i = left;
  j = right + 1;
  pivot = left;   // 先頭要素をpivotとする

  do {
    do { i++; } while (array[i] < array[pivot]);
    do { j--; } while (array[pivot] < array[j]);
    // pivotより小さいものを左へ、大きいものを右へ
    if (i < j) { swap(&array[i], &array[j]); swap2(&kazu[i], &kazu[j]); }
  } while (i < j);

  swap(&array[pivot], &array[j]);   //pivotを更新
  swap2(&kazu[pivot], &kazu[j]);   //pivotを更新

  return j;
}

/* クイックソート */
void quick_sort (long long array[], int left, int right) {
  long long pivot;

  if (left < right) {
    pivot = partition(array, left, right);
    quick_sort(array, left, pivot-1);   // pivotを境に再帰的にクイックソート
    quick_sort(array, pivot+1, right);
  }
}





int main(){
    int n,m;
    long long nedan[100001] = {0};

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

    
    for (int i = 0; i < n; i++) {
       scanf("%lld%d",&nedan[i],&kazu[i]);
    }


    quick_sort(nedan, 0, n-1);




    long long ans = 0;
    int num = 0;

    for (int i = 0; i < m; i++) {
        if (kazu[num] == 0) {
            num++;
        }
        kazu[num]--;
        ans = ans + nedan[num];

    }

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