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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"fmt"
	"sort"
)

type pair struct {
	a int
	b int
}

type pairs []*pair

func (p pairs) Len() int {
	return len(p)
}

func (p pairs) Less(i, j int) bool {
	return p[i].a < p[j].a
}

func (p pairs) Swap(i, j int) {
	p[i], p[j] = p[j], p[i]
}

func main() {
	var N, M int
	fmt.Scan(&N, &M)
	var v pairs
	for i := 0; i < N; i++ {
		var a, b int
		fmt.Scan(&a, &b)
		v = append(v, &pair{a: a, b: b})
	}
	sort.Sort(v)

	sumbuy := 0
	money := 0
	for i := 0; i < N; i++ {
		nokori := M - sumbuy
		buy := min(v[i].b, nokori)
		money += v[i].a * buy
		sumbuy += buy
	}
	fmt.Println(money)
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}