Problem:
Score : 300 points

Problem Statement

A programming competition site AtCode provides algorithmic problems.
Each problem is allocated a score based on its difficulty.
Currently, for each integer i between 1 and D (inclusive), there are p_i problems with a score of 100i points.
These p_1 + … + p_D problems are all of the problems available on AtCode.

A user of AtCode has a value called total score.
The total score of a user is the sum of the following two elements:

Base score: the sum of the scores of all problems solved by the user.

Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of c_i points, aside from the base score (1 ≤ i ≤ D).

Takahashi, who is the new user of AtCode, has not solved any problem.
His objective is to have a total score of G or more points.
At least how many problems does he need to solve for this objective?

Constraints

1 ≤ D ≤ 10

1 ≤ p_i ≤ 100

100 ≤ c_i ≤ 10^6

100 ≤ G

All values in input are integers.

c_i and G are all multiples of 100.

It is possible to have a total score of G or more points.

Input

Input is given from Standard Input in the following format:

D G
p_1 c_1
:
p_D c_D

Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let (D, G): (usize, usize) = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        let mut iter = line.split_whitespace();
        (
            iter.next().unwrap().parse().unwrap(),
            iter.next().unwrap().parse().unwrap()
        )
    };
    let (p, c): (Vec<usize>, Vec<usize>) = {
        let (mut p, mut c) = (vec![], vec![]);
        for _ in 0..D {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            p.push(iter.next().unwrap().parse().unwrap());
            c.push(iter.next().unwrap().parse().unwrap());
        }
        (p, c)
    };

    fn iceil(a: usize, b: usize) -> usize { (a + b - 1) / b }
    let K = 1 << D;
    let ans = (0..K).map(|x| {
        let complete_score = (0..D).map(|i| ((x >> i) & 1) * (100 * (i + 1) * p[i] + c[i])).sum::<usize>();
        let solved = (0..D).map(|i| ((x >> i) & 1) * p[i]).sum::<usize>();
        let (res, _) = (0..D).rev().fold(
            (solved, complete_score),
            |(problem_cnt, score), i| if ((x >> i) & 1) == 0 && G > score {
                let added = std::cmp::min(iceil(G - score, 100 * (i + 1)), p[i]);
                (problem_cnt + added, score + 100 * (i + 1) * added)
            } else { (problem_cnt, score) }
        );
        res
    }).min().unwrap();

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