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)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    token.parse().ok().unwrap()
}

fn main() {
    let d = input::<usize>();
    let g = input::<i64>();
    let (p, c) = (0..d)
        .map(|_| {
            let p = input();
            let c = input();
            (p, c)
        })
        .unzip::<i64, i64, Vec<_>, Vec<_>>();

    println!("{}", dfs(d, 0, 0, d, d, g, &p, &c));
}

fn dfs(
    k: usize,
    count: i64,
    score: i64,
    h: usize,
    d: usize,
    g: i64,
    p: &Vec<i64>,
    c: &Vec<i64>,
) -> i64 {
    if k == 0 {
        if score >= g {
            return count;
        } else if score + (h as i64 * 100 * p[h - 1]) < g {
            return i64::max_value();
        } else {
            let mut c = 0;
            let mut s = score;
            while s < g {
                c += 1;
                s += h as i64 * 100;
            }
            return count + c;
        }
    }

    let mut ans = i64::max_value();
    ans = std::cmp::min(ans, dfs(k - 1, count, score, h, d, g, p, c));
    ans = std::cmp::min(
        ans,
        dfs(
            k - 1,
            count + p[k - 1],
            score + (k as i64 * 100) * p[k - 1] + c[k - 1],
            if k == h { k - 1 } else { h },
            d,
            g,
            p,
            c,
        ),
    );
    ans
}