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::cmp;

macro_rules! read {
    () => {
        {
            let mut line = String::new();
            io::stdin().read_line(&mut line).unwrap();
            line
        }
    };
}

fn split(s: &String) -> std::str::SplitWhitespace {
    s.trim().split_whitespace()
}

fn main() {
    let mut line = read!();
    let mut line = split(&mut line);
    let n: usize = line.next().unwrap().parse().unwrap();
    let m: i64 = line.next().unwrap().parse().unwrap();
    let mut v: Vec<(i64, i64)> = Vec::new();
    for i in 0..n {
        let mut line = read!();
        let mut line = split(&mut line);
        let a: i64 = line.next().unwrap().parse().unwrap();
        let b: i64 = line.next().unwrap().parse().unwrap();
        v.push((a, b));
    }
    v.sort_by_key(|k| k.0);
    let (mut ans, mut cur) = (0, 0);
    for &(a, b) in &v {
        let c = cmp::min(b, m - cur);
        ans += a*c;
        cur += b;
        if cur >= m { break; }
    }
    println!("{}", &ans);
}