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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn main() {
    let input: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let D = input[0];
    let G = input[1];

    let mut ps = Vec::with_capacity(D);
    let mut cs = Vec::with_capacity(D);
    for i in 0..D {
        let input: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
        ps.push(input[0]);
        cs.push(input[1]);
    }

    let mut memo = vec![vec![0; 1010]; 2];

    for i in 0..D {
        let a = (i+1)%2;
        let b = i%2;
        for j in 0..ps[i] {
            for k in j..1010 {
                memo[a][k] = cmp::max(memo[a][k], memo[b][k-j] + (i+1) * j * 100);
            }
        }
        for k in ps[i]..1010 {
            memo[a][k] = cmp::max(memo[a][k], memo[b][k-ps[i]] + cs[i] + ps[i] * (i+1) * 100);
        }
    }

    for i in 0..1010 {
        if memo[D%2][i] >= G {
            println!("{}", i);            
            break;
        }
    }
}