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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>

int main (void){
  int N,M;
  long long int A[10^5];
  int B[20];
  int count=0,sum=0;
  int best_i;
  long long int money=0;
  scanf("%d %d", &N,&M);
  // printf("%d %d\n", N,M);

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

  while(1){
    best_i=0;
    for (int i=1;i<N;i++){
      if(A[best_i]>A[i]){
        best_i=i;
      }
    }
    // printf("best_i=%d\n",best_i);
    if(M<=B[best_i]){ //買占め完了
      money+=A[best_i]*M;
      // printf("LAST buy %lld yen %d hon\n", A[best_i],M);
      break;
    }else{
      M=M-B[best_i];
      money+=A[best_i]*B[best_i];
      // printf("buy %lld yen %d hon\n", A[best_i],B[best_i]);
      A[best_i]=1000000000;
    }
  }

  // printf("money=");
  printf("%lld",money);
  return 0;
}