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::BufRead;

fn main() {
    let stdin = std::io::stdin();
    let mut buf = String::new();
    stdin.read_line(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let _: u32 = iter.next().unwrap().parse().unwrap();
    let g: u32 = iter.next().unwrap().parse().unwrap();

    let mut ps: Vec<u32> = Vec::new();
    let mut cs: Vec<u32> = Vec::new();
    for result in stdin.lock().lines() {
        let line = result.unwrap();
        let mut iter = line.split_whitespace();
        let p: u32 = iter.next().unwrap().parse().unwrap();
        let c: u32 = iter.next().unwrap().parse().unwrap();
        ps.push(p);
        cs.push(c);
    }

    let res = solve(g, &ps, &cs);
    println!("{}", res);
}

fn solve(target: u32, ps: &Vec<u32>, cs: &Vec<u32>) -> u32 {
    let mut min = std::u32::MAX;
    for bit in 0..(1<<ps.len()) {
        let mut sum = 0;
        let mut count = 0;
        for i in 0..ps.len() {
            if bit & (1<<i) != 0 {
                sum += (i as u32 + 1) * 100 * ps[i] + cs[i];
                count += ps[i];
            }
        }
        if sum >= target {
            min = std::cmp::min(min, count);
        } else {
            // 中途半端に解く
            for i in (0..ps.len()).rev() {
                if bit & (1<<i) == 0 {
                    if target > sum + (i as u32 + 1) * 100 * ps[i] {
                        break;
                    } else {
                        let tmp = ((target - sum) as f64 / ((i as u32 + 1) * 100) as f64).ceil() as u32;
                        min = std::cmp::min(min, count + tmp);
                    }
                }
            }
        }
    }
    min
}