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'

var AB;
var buff = [];

function sortData(start,end)
{
	if(start >= end) return;
	var mid = Math.floor((start + end) / 2);
	sortData(start,mid);
	sortData(mid+1,end);
	var p = 0;
	for (i=start; i<=mid; i++) buff[p++] = AB[i];
	var i = mid + 1;
	var j = 0;
	var k = start;
	while ((i<=end) && (j<p))
	{
		if (buff[j] <= AB[i])
		{
			AB[k++] = buff[j++];
		}else{
			AB[k++] = AB[i++];
		}
	}
	while (j < p) AB[k++] = buff[j++];
}

function Main(input){
	let temp = input.trim().split("\n");
	let NM = temp[0].split(" ").map( (v)=>(parseInt(v)) );
	AB = new Array(NM[0]);
	buff = new Array(NM[0]);
	for(let i=0;i<NM[0];i++){
		AB[i] = temp[i+1].split(" ").map( (v)=>(parseInt(v)) );
	}
	//AB.sort((a,b)=>(a[0] < b[0])?-1:1);
	sortData(0,AB.length-1);

	let count = NM[1];
	let money = 0;
	for(let i=0;i<NM[0];i++){
		if(AB[i][1] < count){
			//全部加算
			money += AB[i][0] * AB[i][1];
			count -= AB[i][1];
		} else {
			money += AB[i][0] * count;
			break;
		}
	}
	console.log(money);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));