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>
//小さい順にソート
long long c[100002][2];
void merge(int i,int j,long long s[][2]){
  if(i!=j){
    merge(i,(i+j)/2,s);
    merge((i+j)/2+1,j,s);
    int a=i,b=(i+j)/2+1;
    int d;
    for(d=i;d<=j;d++){
      if((i+j)/2<a){
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
      else if(j<b){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else if(s[a][0]<s[b][0]){
        c[d][0]=s[a][0];
        c[d][1]=s[a][1];
        a++;
      }
      else{
        c[d][0]=s[b][0];
        c[d][1]=s[b][1];
        b++;
      }
    }
    for(d=i;d<=j;d++){
      s[d][0]=c[d][0];
      s[d][1]=c[d][1];
    }
  }
}

int main(){
  int n,i;
  long long m;
  scanf("%d%lld",&n,&m);
  long long a[n+1][2];
  for(i=0;i<n;i++)
    scanf("%lld%lld",&a[i][0],&a[i][1]);
  merge(0,n-1,a);
  long long count=0,money=0;
  i=0;
  while(count!=m){
    if(count+a[i][1]<m){
      count+=a[i][1];
      money+=a[i][0]*a[i][1];
      i++;
    }
    else{
      money+=a[i][0]*(m-count);
      count=m;
    }
  }
  printf("%lld\n",money);
  return 0;
}