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::iter;
use std::u32;

fn main(){
    let d: u32 = read_one();
    let g: u32 = read_one();
    let pc: Vec<(u32, u32)> = read2_lines(d as usize);
    
    let mut v: Vec<u32> = Vec::new();
    
    for i in 0..(u32::pow(2, d)) {
        let (mut s, mut n) = pc.iter().enumerate()
            .filter(|&(index, _item)| {((i >> index) & 1) == 1})
            .map(|(index, &(p, c))| {(p * (index + 1) as u32 * 100 + c, p)})
            .fold((0, 0), |x, y| {(x.0 + y.0, x.1 + y.1)});
        for (index, &(p, c)) in pc.iter()
            .enumerate()
            .filter(|&(index, _item)| {((i >> index) & 1) != 1})
            .rev() 
        {
            if g <= s {break;}
            let need = (g - s - 1) / ((index + 1) as u32 * 100) + 1;
            if need <= p - 1 {
                n += need;
                s += need * (index + 1) as u32 * 100;
                break;
            } else {
                n += p - 1;
                s += (p - 1) * (index + 1) as u32 * 100;
            }
        }
        if g <= s {
            v.push(n);
        }
    }
    println!("{}", v.iter().min().unwrap())
}

use std::io::*;
use std::str::*;

fn read_one<T: std::str::FromStr>() -> T {
    let mut s = stdin().bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.trim().parse().ok().unwrap()
}

fn read2<T: FromStr, T2: FromStr>() -> (T, T2) {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    let mut iter = s.trim().split_whitespace();
    (iter.next().unwrap().parse().ok().unwrap(), iter.next().unwrap().parse().ok().unwrap())
}

fn read2_lines<T: FromStr, T2: FromStr>(n: usize) -> Vec<(T, T2)> {
    (0..n).map(|_| read2()).collect()
}