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)



#[allow(unused_macros)]
macro_rules! dbg {
    ($val:expr) => {
        match $val {
            tmp => {
                eprintln!("[{}:{}] {} = {:#?}",
                          file!(), line!(), stringify!($val), &tmp);
                tmp
            }
        }
    }
}

#[allow(dead_code)]
pub struct Scanner<'a> {
    _buf: &'a str,
    iter: ::std::str::SplitWhitespace<'a>,
}

#[allow(dead_code)]
impl<'a> Scanner<'a> {
    pub fn new(buf: &'a str) -> Self {
        Scanner {
            _buf: buf,
            iter: buf.split_whitespace(),
        }
    }

    pub fn array<T>(&mut self, n: usize) -> Vec<T>
        where T: ::std::str::FromStr,
              T::Err: ::std::fmt::Debug,
    {
        let mut res = Vec::with_capacity(n);
        for _ in 0..n {
            res.push(self.value());
        }
        res
    }

    pub fn value<T>(&mut self) -> T
        where T: ::std::str::FromStr,
              T::Err: ::std::fmt::Debug,
    {
        self.next().parse().unwrap()
    }

    pub fn chars(&mut self) -> Vec<char> {
        self.next().chars().collect()
    }

    fn next(&mut self) -> &str {
        self.iter.next().unwrap()
    }
}

fn div_ceil(a: i64, b: i64) -> i64 {
    (a+b-1) / b
}

fn f(d: i64, mut g: i64, p: &[i64], c: &[i64], mask: i64) -> i64 {
    let mut res = 0;

    for i in 0..d {
        if mask & (1<<i) != 0 {
            g -= p[i as usize] * 100*(i+1);
            g -= c[i as usize];
            res += p[i as usize];
        }
    }
    if g <= 0 { return res; }

    for i in (0..d).rev() {
        if mask & (1<<i) != 0 { continue; }

        let q = div_ceil(g, 100*(i+1));
        if q <= p[i as usize] - 1 {
            g -= q * (100*(i+1));
            res += q;
            break;
        }
        else {
            g -= (p[i as usize] - 1) * (100*(i+1));
            res += p[i as usize] - 1;
        }
    }
    if g > 0 { return 10_i64.pow(18); }

    res
}

fn main() {
    let mut _input = String::new();
    {
        use ::std::io::Read;
        ::std::io::stdin().read_to_string(&mut _input).unwrap();
    }
    let mut sc = Scanner::new(&_input);

    let d = sc.value::<i64>();
    let g = sc.value::<i64>();
    let mut p = Vec::with_capacity(d as usize);
    let mut c = Vec::with_capacity(d as usize);
    for _ in 0..d {
        p.push(sc.value::<i64>());
        c.push(sc.value::<i64>());
    }

    let mut ans = 10_i64.pow(18);
    for i in 0..(1<<d) {
        let cur = f(d, g, &p, &c, i);
        if cur < ans {
            ans = cur;
        }
    }

    println!("{}", ans);
}