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>


  /* 配列の要素を交換する */
void Swap(int x[ ], int i, int j)
{
    int temp;

    temp = x[i];
    x[i] = x[j];
    x[j] = temp;
}
/* クイックソートを行う */
void QSort(int x[ ], int y[], int left, int right)
{
    int i, j;
    int pivot;

    i = left;                      /* ソートする配列の一番小さい要素の添字 */
    j = right;                     /* ソートする配列の一番大きい要素の添字 */

    pivot = x[(left + right) / 2]; /* 基準値を配列の中央付近にとる */

    while (1) {                    /* 無限ループ */

        while (x[i] < pivot)       /* pivot より大きい値が */
            i++;                   /* 出るまで i を増加させる */

        while (pivot < x[j])       /* pivot より小さい値が */
            j--;                   /*  出るまで j を減少させる */
        if (i >= j)                /* i >= j なら */
            break;                 /* 無限ループから抜ける */

        Swap(x, i, j);             /* x[i] と x[j]を交換 */
        Swap(y, i, j);
        i++;                       /* 次のデータ */
        j--;
    }
    if (left < i - 1)              /* 基準値の左に 2 以上要素があれば */
        QSort(x, y,left, i - 1);     /* 左の配列を Q ソートする */
    if (j + 1 <  right)            /* 基準値の右に 2 以上要素があれば */
        QSort(x, y,j + 1, right);    /* 右の配列を Q ソートする */
}



int main() {
  int n,m;
  scanf("%d %d", &n, &m);
  int a[n];
  int b[n];
  int i,j;
  long tmp,sum,money;
  for(i=0;i<n;i++){
    scanf("%d %d", &a[i], &b[i]);
  }
  
  QSort(a, b, 0, n-1);
  sum = 0;
  money = 0;
  for (i=0; i<n; ++i) {
    if(m >= sum+b[i]){
      money += a[i]*b[i];
      sum += b[i];
    }
    else{
      money += (m-sum)*a[i];
      break;
    }
  }

  printf("%ld", money);
  return 0;
}