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: Wrong Answer)

#[allow(unused_imports)]
use std::cmp::{min,max};
#[allow(unused_imports)]
use std::collections::BTreeMap;
#[allow(unused_imports)]
use std::ops::*;
#[allow(unused_imports)]
use std::collections::BinaryHeap;

fn main() {
    let (d,g) = {
        let xs = read_vec_i64();
        (xs[0] as usize, xs[1] as usize)
    };
    let (ps, cs) = {
        let mut ps = vec![];
        let mut cs = vec![];
        for _ in 0..d {
            let mut xs = read_vec_i64();
            ps.push(xs[0] as usize);
            cs.push(xs[1] as usize / 100);
        }
        (ps,cs)
    };
    let inf = 1usize << 40;
    let mut xs = vec![inf; 14234];
    let mut ys = vec![inf; 14234];
    let m = 1001;
    xs[0] = 0;
    for i in 0..d {
        for k in 0..m {
            ys[k] = xs[k];
        }
        for k in 0..m {
            for j in 1..ps[i] {
                let t = k+(i+1)*(j);
                ys[k+(i+1)*(j)] = min(ys[k+(i+1)*(j)], xs[k] + j);
            }
            ys[k+(i+1)*ps[i]+cs[i]] = min(ys[k+(i+1)*ps[i]+cs[i]], xs[k] + ps[i]);
        }
        for k in 0..m {
            xs[k] = ys[k];
        }
    }
    let mut res = inf;
    for i in g/100..14234 {
        res = min(res, xs[i]);
    }
    println!("{}", res);
}

#[allow(dead_code)]
fn read_line() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok();
    ret.pop();
    return ret;
}

#[allow(dead_code)]
fn read_i64() -> i64 {
    let ss = read_line();
    return ss.parse::<i64>().unwrap();
}

#[allow(dead_code)]
fn read_vec_i64() -> Vec<i64> {
    let mut res = vec![];
    let ss = read_line();
    for ts in ss.split_whitespace() {
        let x = ts.parse::<i64>().unwrap();
        res.push(x);
    }
    return res;
}

use std::fmt::Display;
#[allow(dead_code)]
fn write_vec<T: Display>(xs: &Vec<T>) {
    if xs.len() == 0 {
        println!("");
        return;
    }
    print!("{}", xs[0]);
    for i in 1..xs.len() {
        print!(" {}", xs[i]);
    }
    println!("");
}

#[allow(dead_code)]
fn tf<T>(cond: bool, t: T, f: T) -> T {
    if cond { t } else { f }
}