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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

int main()
{
 
//  int a, b, c, d, e, X, Y;
  int a, b;
  //  double  c;
  long long  A[1000001]={};
  long long  B[1000001]={};
//  int  B[100003]={};
  int  L[1003]={};

  int  flag[10000003]={};

//  long long  a, b, c, d, X, Y;
  long long  c, d, X, Y;
char s[100001]={};
  long long JPY;
  double BTC;
char t[100001]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
//  int flag=0;
  int tmp=0;
//  int min=1000000;
  long long min=10000000001;
  int len=0;
  int cnt=0;
  int cnt0=0;
  int cnt1=0;
  int num=0;
//  double ans=0.0;
  long long ans=0;
  int max=0;
  int sum=0;
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
//  scanf("%lld %lld", &a, &b);
  scanf("%d%d", &a,&b);
  for(i=0;i<a;i++){
    scanf("%lld%lld", &A[i],&B[i]);
  }
  while(b+1>cnt){
    min=10000000001;
    num=0;
    for(i=0;i<a;i++){
      //min
      if((min>A[i]) && (flag[i]==0)){
        min=A[i];
        num=i;
      }
    }
    flag[num]=1;
    if(b<(cnt+B[num])){
      for(i=0;i<(B[num]+1);i++){
//        printf("cnt%d  ", cnt);

        if(b<=cnt){
          ans=ans+(A[num]*tmp);
          printf("%lld\n", ans);
          return 0;
        }
        else {
          cnt++;
          tmp++;
        }
      }
    }
    else{
      ans=ans+(A[num]*B[num]);
      cnt=cnt+B[num];
//      printf("%lld num%d\n", ans, num);
    }
//    printf("cnt%d  ", cnt);
  }
  



  //    if(max < cnt - max)  printf("Yes\n");
//    else  
  
return 0;
}