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)

macro_rules! input {
    ($t : ty) => {{
        use std::io::BufRead;
        let tmp_handle = std::io::stdin();
        let mut handle = tmp_handle.lock();
        let mut buf = String::new();
        handle.read_line(&mut buf).unwrap();
        let tmp_var: $t = buf.trim().parse().unwrap();
        buf.clear();
        tmp_var
    }};
    ($t : ty, $size : expr) => {{
        use std::io::BufRead;
        let tmp_handle = std::io::stdin();
        let mut handle = tmp_handle.lock();
        let mut buf = String::new();
        handle.read_line(&mut buf).unwrap();
        let tmp_var: $t = buf
            .trim()
            .split_whitespace()
            .map(|a| a.parse().unwrap())
            .collect();
        buf.clear();
        tmp_var
    }};
}
fn bit_patterns(digit: usize) -> Vec<Vec<usize>> {
    let until = usize::pow(2, digit as u32);
    let mut v = Vec::with_capacity(until);
    for num in 0..until {
        v.push(bits(num, digit));
    }
    return v;
}
fn bits(mut num: usize, digit: usize) -> Vec<usize> {
    let mut v = Vec::with_capacity(digit);
    for _ in 0..digit {
        v.push(num % 2);
        num /= 2;
    }
    return v;
}
use std::cmp::min;
use std::cmp::Ordering;
#[derive(Debug)]
struct SomeType {
    value: usize,
    info: (usize, usize, usize),
}
impl Eq for SomeType {}
impl PartialEq for SomeType {
    fn eq(&self, other: &Self) -> bool {
        self.value == other.value
    }
}
impl PartialOrd for SomeType {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        self.value.partial_cmp(&other.value)
    }
}
impl Ord for SomeType {
    fn cmp(&self, other: &Self) -> Ordering {
        self.value.cmp(&other.value)
    }
}

fn main() {
    let config = input!(Vec<usize>, 2);
    let mut questions = Vec::with_capacity(config[0]);
    for i in 0..config[0] {
        let question = input!(Vec<usize>, 2);
        questions.push(SomeType {
            value: question[0] * (i + 1) * 100 + question[1],
            info: (i, question[0], question[1]),
        });
    }
    let all_bit: Vec<Vec<usize>> = bit_patterns(config[0]);
    let mut min_q_count: isize = std::isize::MAX;
    for bit in all_bit {
        let mut q_count = 0;
        let mut total: isize = 0;
        for (i, b) in bit.iter().enumerate() {
            if *b == 1 {
                total += questions[i].value as isize;
                q_count += questions[i].info.1 as isize;
            } 
        }
        if total < config[1] as isize { continue; }

        // delete exceed point 
        let turning_question = SomeType { value: 0, info: (0, 0, 0)};
        let mut tmp_min_q_count = q_count;
        for (i, q) in bit.iter().enumerate().filter(|a| *a.1 == 1) {
            let mut tmp_q_count = q_count;
            let mut exceed_points = total - config[1] as isize;
            if exceed_points - questions[i].info.2 as isize - ((questions[i].info.0)+1) as isize * 100  >= 0 {
                exceed_points -= (questions[i].info.2 as isize + ((questions[i].info.0)+1) as isize * 100); 
                tmp_q_count -= 1;
                // neadless solve questions
                let nsq = exceed_points / ((i+1) as isize * 100);
                tmp_q_count -= min(nsq, (questions[i].info.1 -1) as isize);
            }
            if tmp_min_q_count > tmp_q_count { tmp_min_q_count = tmp_q_count; }
        }
        q_count = tmp_min_q_count;
        if q_count < min_q_count { min_q_count = q_count; }
    }
    println!("{}", min_q_count);
}