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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let m: i64 = iter.next().unwrap().parse().unwrap();
    let mut a: Vec<i64> = vec![];
    let mut b: Vec<i64> = vec![];

    for _ in 0..n {
        a.push(iter.next().unwrap().parse().unwrap());
        b.push(iter.next().unwrap().parse().unwrap());
    }

    let mut p: Vec<usize> = (0..n).collect();
    p.sort_by(|p,q| a[*p].cmp(&a[*q]));
    let mut ans = 0_i64;
    let mut buyed = 0;
    for i in p {
        //println!("{} {}", a[i], b[i]);
        let buy = std::cmp::min(m - buyed, b[i]);
        if buy == 0 { break; }
        buyed += buy;
        ans += buy * a[i];
    }
    println!("{}", ans);
}