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: Wrong Answer)

use std::io::Read;

fn main() {
    let mut buf = String::new();

    // 標準入力から全部bufに読み込む
    std::io::stdin().read_to_string(&mut buf).unwrap();

    // 行ごとのiterが取れる
    let mut iter = buf.split_whitespace();
    let problem_counts: usize = iter.next().unwrap().parse().unwrap();
    let target_score: usize = iter.next().unwrap().parse().unwrap();
    let mut bornases: Vec<(usize, usize, usize)> = (0..problem_counts)
        .enumerate()
        .map(|(i, _)| {
            (
                (i + 1) * 100,
                iter.next().unwrap().parse::<usize>().unwrap(),
                iter.next().unwrap().parse::<usize>().unwrap(),
            )
        })
        .map(|x| (x.0, x.1, x.0 * x.1 + x.2))
        .collect();

    #[derive(Clone, Debug)]
    struct Pattern {
        attempts: usize,
        score: usize,
        another_list: Vec<(usize, usize)>,
    }

    // 各行を選ぶ選ばないパターンを作る
    let mut patterns = Vec::new();
    // 2の3乗の組み合わせを作成
    for i in 0..2usize.pow(problem_counts as u32) {
        let mut row = Pattern {
            attempts: 0,
            score: 0,
            another_list: vec![(
                bornases[problem_counts - 1].0,
                bornases[problem_counts - 1].1,
            )],
        };
        // ビット演算する際の最大桁が1桁目に来るとこで
        // シフトできれば良いので、number.len()-1までloop
        let anothers: Vec<(usize, usize)> = (0..problem_counts)
            .filter(|j| ((i >> j) & 1) != 1)
            .map(|j| (bornases[j].0, bornases[j].1))
            .collect();
        for j in 0..problem_counts {
            // j桁右シフトして最初のbitが1かチェック
            if ((i >> j) & 1) == 1 {
                row = Pattern {
                    attempts: row.attempts + bornases[j].1,
                    score: row.score + bornases[j].2,
                    another_list: anothers.clone(),
                };
            }
        }
        patterns.push(row);
    }
    patterns = patterns[0..patterns.len() - 1].to_vec();

    let mut result = 10000000;
    for pattern in &patterns {
        let mut tmp_count = 0;
        if pattern.score >= target_score {
            tmp_count = pattern.attempts;
        } else {
            let mut current_score = pattern.score;
            tmp_count = pattern.attempts;
            'outer: for anothers in pattern.another_list.iter().rev() {
                for _ in 0..anothers.1 - 1 {
                    if current_score < target_score {
                        tmp_count += 1;
                        current_score += anothers.0;
                    } else {
                        break 'outer;
                    }
                }
            }
            if current_score < target_score {
                continue;
            }
        }
        if result > tmp_count && tmp_count != 0 {
            result = tmp_count;
        }
    }

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