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, dead_code)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros, dead_code)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros, dead_code)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(dead_code)]
struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
    size: Vec<usize>,
}

#[allow(dead_code)]
impl UnionFind {
    fn new(n: usize) -> UnionFind {
        let mut p = vec![0; n];
        for i in 0..n {
            p[i] = i;
        }
        return UnionFind {
            parent: p,
            rank: vec![0; n],
            size: vec![1; n],
        };
    }

    fn find(&mut self, x: usize) -> usize {
        if x == self.parent[x] {
            x
        } else {
            let p = self.parent[x];
            let pr = self.find(p);
            self.parent[x] = pr;
            pr
        }
    }

    fn same(&mut self, a: usize, b: usize) -> bool {
        self.find(a) == self.find(b)
    }

    fn unite(&mut self, a: usize, b: usize) {
        let a_root = self.find(a);
        let b_root = self.find(b);
        if a_root == b_root {
            return;
        }
        if self.rank[a_root] > self.rank[b_root] {
            self.parent[b_root] = a_root;
            self.size[a_root] += self.size[b_root];
        } else {
            self.parent[a_root] = b_root;
            self.size[b_root] += self.size[a_root];
            if self.rank[a_root] == self.rank[b_root] {
                self.rank[b_root] += 1;
            }
        }
    }

    fn get_size(&mut self, x: usize) -> usize {
        let root = self.find(x);
        self.size[root]
    }
}

const MOD_P: usize = 1000000007;

#[allow(dead_code)]
// fact(n) = n! mod p
fn fact(n: usize) -> usize {
    let mut acc = 1;
    for i in 1..n + 1 {
        acc = acc * i % MOD_P;
    }
    acc
}

#[allow(dead_code)]
fn mod_pow(b: usize, mut e: usize) -> usize {
    let mut base = b;
    let mut acc = 1;

    while e > 1 {
        if e % 2 == 1 {
            acc = acc * base % MOD_P;
        }
        e /= 2;
        base = base * base % MOD_P;
    }

    if e == 1 {
        acc = acc * base % MOD_P;
    }

    acc
}

#[allow(dead_code)]
fn comb(n: usize, r: usize) -> usize {
    // nCr = n! / (r! (n-r)!) = n! (r!)^(p-2) ((n-r)!)^(p-2)
    let x = ((n - r + 1)..(n + 1)).fold(1, |p, x| p * x % MOD_P);
    let y = mod_pow(fact(r), MOD_P - 2);
    x * y % MOD_P
}

#[allow(dead_code)]
fn gcd(a: usize, b: usize) -> usize {
    if b > a {
        gcd(b, a)
    } else if a % b == 0 {
        b
    } else {
        gcd(b, a % b)
    }
}

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

// use std::cmp::{max, min};

fn main() {
    input! {
        d: usize,
        g: usize,
        pc: [(usize,usize);d],
    }

    let mut ans = std::usize::MAX;

    for allsolve in 0..(1 << d) {
        let mut points = 0;
        let mut problems = 0;
        for i in 0..d {
            if ((1 << i) & allsolve) != 0 {
                let (p, c) = pc[i];
                points += 100 * (i + 1) * p + c;
                problems += p;
            }
        }

        if points >= g {
            ans = std::cmp::min(problems, ans);
            continue;
        }

        for i in 0..d {
            if ((1 << i) & allsolve) == 0 {
                let (p, c) = pc[i];
                let rest = g - points;
                if rest <= (100 * (i + 1) * p) {
                    let probs = (rest + (100 * (i + 1) - 1)) / (100 * (i + 1));
                    ans = std::cmp::min(ans, problems + probs);
                }
            }
        }
    }
    println!("{}", ans);
}