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)

fn main() {
    use std::io::Read;
    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::<usize>().unwrap();
    let m: u64 = iter.next().unwrap().parse::<u64>().unwrap();
    let mut pair: Vec<(u64, u64)> = Vec::with_capacity(n);
    
    for _ in 0..n {
        pair.push((
            iter.next().unwrap().parse::<u64>().unwrap(),
            iter.next().unwrap().parse::<u64>().unwrap()
        ));
    }

    pair.sort();
    let mut money = 0u64;
    let mut num = 0u64;
    for i in 0..n {
        if num + pair[i].1 < m {
            money += pair[i].0 * pair[i].1;
            num += pair[i].1;
        } else {
            let k = m - num;
            money += pair[i].0 * k;
            break;
        }
    }
    println!("{}", money);
}