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;
use std::str::FromStr;

#[allow(dead_code)]
fn readval<T: FromStr>() -> T {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    
    buf.trim().parse::<T>().ok().unwrap()
}

fn readvec<T: FromStr>() -> Vec<T> {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();

    buf.trim().split(" ")
              .map(|c| c.parse::<T>().ok().unwrap())
              .collect::<Vec<T>>()
}

fn main() {
    let nm = readvec::<usize>();
    let (n, m) = (nm[0], nm[1]);

    let mut ab: Vec<Vec<u64>> = vec![];
    for _ in 0..n {
        ab.push(readvec::<u64>());
    }
    
    ab.sort();

    let mut res = m as u64;
    let mut ans = 0;

    for i in 0..n {
        if res == 0 {
            break;
        }

        else if res >= ab[i][1] {
            ans += ab[i][0] * ab[i][1];
            res -= ab[i][1];
        }

        else {
            ans += ab[i][0] * res;
            res = 0;
        }
    }

    println!("{:?}", ans);


}