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 (1.42.0) (judge verdict: Accepted)

use proconio::input;
use std::cmp;

fn main() {
    input! {
        d: usize,
        g: isize,
        pro: [(isize, isize); d],
    }
    let mut min = std::isize::MAX;

    for bit in 0..1 << d {
        let mut score = 0;
        let mut count = 0;
        let mut problem = pro.clone();
        for i in 0..d {
            if (bit >> i) & 1 != 0 {
                score += 100 * (i as isize + 1) * problem[i].0 + problem[i].1;
                count += pro[i].0;
                problem[i].0 = 0;
            }
        }
        if score >= g {
            min = cmp::min(min, count);
            continue;
        }
        let mut remain = g - score;
        //println!("bit {} remain {}, count {}", bit, remain, count);
        //greedy
        for i in (0..d).rev() {
            if problem[i].0 == 0 {
                continue;
            }
            if remain <= 0 {
                min = cmp::min(min, count);
                break;
            }
            let diff = 100 * (i as isize + 1);
            let want = (remain + (diff - 1)) / diff;
            if want > problem[i].0 {
                remain -= diff * problem[i].0 + problem[i].1;
                count += problem[i].0;
                problem[i].0 = 0;
            } else {
                remain -= diff * want;
                count += want;
            }
            //println!("bit {} remain {} count {}", bit, remain, count);
        }
        min = cmp::min(min, count);
    }
    println!("{}", min);
}