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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

'use strict'
function Main(input){
  input=input.trim().split(/\s+/).map(x=>x-0);
  var N=input.shift();
  var M=input.shift();
  var A=[];
  var B=[];
  for(let i=0;i<N;i++){
    A[i]=input[2*i];
    B[i]=input[2*i+1];
  }
   A=binarySearchSort(A);
//  console.log(A);
  var accumulatedCost=0;
  var i=0;
   while(M>=B[A[i][1]]){
     M-=B[A[i][1]];
     accumulatedCost+=A[i][0]*B[A[i][1]];
     i++;
   }
  if(M>0)accumulatedCost+=A[i][0]*M;
  accumulatedCost=accumulatedCost+'';
  console.log(accumulatedCost);
}

function binarySearchSort(arr){
  var N=arr.length;
  var newArr=[[arr[0],0]];
  for(let i=0;i<N;i++){
    let n=newArr.length;
    var d=0,m=0,u=n-1;
    m=Math.floor((d+u)/2);
    while(!(d>u)){
      if(newArr[m][0]<arr[i]){
        d=m+1;
      }else{
        u=m-1;
      }
      m=Math.floor((d+u)/2);
    }
    newArr.splice(m+1,0,[arr[i],i]);
  }
  return newArr;
}

Main(require("fs").readFileSync('/dev/stdin','utf8'));