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>
int main(){
  int N,M;
  int A[100000][2];
  scanf("%d %d",&N,&M);
  int a,b,i,j,k,l;
  
  for (i=0;i<N;i++){
    scanf("%d %d",&a,&b);
    if (i==0){
      A[0][0] = a;
      A[0][1] = b;
    }else{
      for (j=0;j<i;j++){
        if(a < A[j][0]){
          for (k=i-1;k>=j;k--){
            A[k+1][0] = A[k][0];
            A[k+1][1] = A[k][1];
            A[j][0] = a;
            A[j][1] = b;
          }
          j=i+1;
        }
        if(j=i-1 && a >= A[j][0]){
          A[i][0] = a;
          A[i][1] = b;
        }
      }
    }
  }
  int sum = 0;
  int money = 0;
  for (l=0;l<N;l++){
    if (sum + A[l][1] > M){
      money += A[l][0]*(M - sum);
      l = N+1;
    }else if(sum + A[l][1] <= M){
      money += A[l][0]*A[l][1];
    }
  }
  printf("%d\n",money);
  return 0;
}