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::borrow::Borrow;
use std::cmp::{max, min};
use std::io::{stdin, Read};
use std::str::FromStr;

// https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

#[allow(dead_code)]
fn main() {
    let d = read::<usize>();
    let g = read::<usize>();
    let n = d;
    let pcs = (0..n)
        .map(|_| (read::<usize>(), read::<usize>()))
        .collect::<Vec<_>>();
    let pcs = pcs.into_iter().enumerate().collect::<Vec<_>>();

    let width = 1 << n;
    let t = (0..width).map(|bits| (0..n).map(move |i| bits >> i & 1));
    let mut ans = usize::max_value();

    for x in t {
        let fil = x.collect::<Vec<_>>();
        let total_value: usize = pcs
            .iter()
            .map(|&(i, (p, c))| {
                if fil[i] == 1 {
                    (i + 1) * 100 * p + c
                } else {
                    0
                }
            })
            .sum();
        let total_count: usize = pcs
            .iter()
            .map(|&(i, (p, c))| if fil[i] == 1 { p } else { 0 })
            .sum();
        let rest = (&pcs).into_iter().filter(|x| fil[x.0] == 0);

        if total_value >= g {
            ans = min(ans, total_count)
        } else {
            let mut total_count = total_count;
            let mut total_value = total_value;

            'for1: for &(i, (p, c)) in rest.rev() {
                let div = if (g - total_value) % ((i + 1) * 100) == 0 {
                    (g - total_value) / ((i + 1) * 100)
                } else {
                    (g - total_value) / ((i + 1) * 100) + 1
                };
                let pp = min(div, p);
                total_count += pp;
                total_value += pp * ((i + 1) * 100);

                if total_value >= g {
                    ans = min(ans, total_count);
                    break 'for1;
                }
            }
        }
    }
    println!("{}", ans);
}