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: Runtime Error)

extern crate core;

use std::io::{self, Read};

#[macro_use]
mod parser {
    macro_rules! input {
    ($s:expr=>$($t:tt)*) => {
        let mut lines=$s.split("\n");
        $(
            line_parse!(lines,$t);
        )*
    };
    }

    macro_rules! line_parse {
    ($lines:expr,($($name:ident:$t:tt)*)) => {
        let mut line=$lines.next().unwrap().split_whitespace();
        $(value_def!(line,$name,$t);)*
    };

    //複数行
    ($lines:expr,{$n:expr;$name:ident:$t:tt}) => {
        values_def!($lines,$n,$name,$t);
    };
    }

    macro_rules! value_def {
        ($line:expr, $name:ident, $t:tt) => {
            let $name = value!($line, $t);
        };
    }

    macro_rules! values_def {
        ($lines:expr, $n:expr, $name:ident, $t:tt) => {
            let $name = {
                let mut vec = Vec::new();
                for i in 0..$n {
                    let mut next = $lines.next().unwrap().split_whitespace();
                    vec.push(value!(next, $t));
                }
                vec
            };
        };
    }

    macro_rules! value {
    //配列
    ($line:expr,[$t:tt]) => {
        $line.map(|x|{
        let mut iter=::std::iter::once(x);
        value!(iter,$t)
        }).collect::<Vec<_>>()
    };
    //タプル
    ($line:expr,($($t:tt),*)) => {
        ($(value!($line,$t),)*)
    };
    //文字列
    ($line:expr,#) => {
        $line.next().unwrap()
    };
    //単一値
    ($line:expr,$t:ty) => {
        $line.next().unwrap().parse::<$t>().unwrap()
    };
    }
}

fn main() {
    let mut input = String::new();
    io::stdin().read_to_string(&mut input).unwrap();
    let output = solve(input.trim().to_string());
    println!("{}", output);
}

fn solve(input: String) -> String {
    input!(input=>(s:#));
    let mut list = s.chars().collect::<Vec<_>>();
    list.reverse();

    let q_count = list.clone().into_iter().filter(|&x| x == '?').count();
    let mut q_count_pow_mod = 1;
    for _ in 0..q_count {
        q_count_pow_mod = mul(q_count_pow_mod, 3);
    }

    let mut a = 0i64;
    let mut b = 0i64;
    let mut c = 0i64;
    for x in list {
        match x {
            'A' => a = add(a, b),
            'B' => b = add(b, c),
            'C' => c = add(c, q_count_pow_mod),
            '?' => {
                a = add(a, div(b, 3));
                b = add(b, div(c, 3));
                c = add(c, div(q_count_pow_mod, 3));
            }
            _ => unreachable!(),
        }
    }
    a.to_string()
}

fn power(x: i64, y: i64) -> i64 {
    if y == 0 {
        1
    } else if y == 1 {
        x % 1000000007
    } else if y % 2 == 0 {
        power(x, y / 2).pow(2) % 1000000007
    } else {
        power(x, y / 2).pow(2) * x % 1000000007
    }
}

fn div(a: i64, b: i64) -> i64 {
    mul(a, power(b, 1000000007 - 2))
}

fn add(a: i64, b: i64) -> i64 {
    (a + b) % 1000000007
}

fn mul(a: i64, b: i64) -> i64 {
    ((a % 1000000007) * (b % 1000000007)) % 1000000007
}

macro_rules! tests {
    ($($name:ident: $input:expr=>$output:expr,)*) => {
        mod tests {
            $(
                #[test]
                fn $name() {
                    assert_eq!($output.trim().to_string(),super::solve($input.trim().to_string()));
                }
            )*
        }
    }
}

tests! {
    test1: "A??C" => "8",
    test2: "ABCBC" => "3",
    test3: "????C?????B??????A???????" => "979596887",
    test_add1:"A?C"=>"1",
}