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::str::FromStr;
use std::fmt::Debug;
use std::cmp;

#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()

}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn list_patterns(n: usize, v: &mut Vec<bool>) -> Vec<Vec<bool>>{

    if v.len() == n{
        return vec![v.clone()];
    }

    let mut res: Vec<Vec<bool>> = vec![];
    v.push(false);
    let mut tmp = list_patterns(n, v);
    res.append(&mut tmp);
    v.pop();

    v.push(true);
    let mut tmp = list_patterns(n, v);
    res.append(&mut tmp);
    v.pop();

    return res;
}

fn main(){
    let line = read_as_vec::<usize>();
    let (d, g) = (line[0], line[1]);

    let mut problems = vec![];
    for i in 0..d{
        let line = read_as_vec::<usize>();
        problems.push((line[0], line[1], (i+1)*100));
    }

    problems.sort_by_key(|x| x.1);
    problems.reverse();
    let mut v = vec![];
    let patterns = list_patterns(d, &mut v);

    let mut ans = std::usize::MAX;

    for pat in patterns{
        //println!("{:?}", pat);
        let completed_sum: usize = pat.iter().zip(problems.iter()).filter(|x| *x.0)
            .map(|x| (x.1).2*(x.1).0 + (x.1).1).sum::<usize>();
        let completed_num: usize = pat.iter().zip(problems.iter()).filter(|x| *x.0)
            .map(|x| (x.1).0).sum::<usize>();

        if completed_sum >= g{
            ans = cmp::min(ans, completed_num);
            continue;
        }
        //.println!("complete_sum = {}, complete_num={}", completed_sum, completed_num);
        let (num, _, points) = *pat.iter().zip(problems.iter()).
            filter(|x| *x.0 == false).map(|x| x.1).nth(0).unwrap();
        //println!("points ={}, num = {}, g = {}", points, num, g);
        let solved = (g - completed_sum+points-1)/points;
        //println!("solved = {}", solved);
        if solved <= num{
            ans = cmp::min(ans, completed_num + solved);
        }
        //println!("ans = {}", ans);
    }

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