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_mut)]
#![allow(non_snake_case)]
#![allow(unused_macros)]
#![allow(unused_variables)]
#![allow(unused_imports)]
#![allow(dead_code)]
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
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)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

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

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>>()
    };

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

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

macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => {
        cmp::max($x, max!($($y),+))
    };
}

macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => {
        cmp::min($x, min!($($y),+))
    };
}

const MOD: usize = 1_000_000_007;
const MAX: usize = 510_000;

struct BinCoeff {
    fac: Vec<usize>,
    finv: Vec<usize>,
    inv: Vec<usize>,
}

impl BinCoeff {
    fn new() -> BinCoeff {
        let fac = vec![0; MAX];
        let finv = vec![0; MAX];
        let inv = vec![0; MAX];
        let mut bincoeff = BinCoeff { fac: fac, finv: finv, inv: inv };
        bincoeff.com_init();
        bincoeff
    }

    fn com_init(&mut self) {
        let fac = &mut self.fac;
        let finv = &mut self.finv;
        let inv = &mut self.inv;
        fac[0] = 1;
        fac[1] = 1;
        finv[0] = 1;
        finv[1] = 1;
        inv[1] = 1;
        for i in 2..MAX {
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    fn bin_coeff(&mut self, n: usize, k: usize) -> usize {
        if k > n/2 {
            self.bin_coeff(n-k, k)
        } else {
            self.fac[n] * (self.finv[k] * self.finv[n - k] % MOD) % MOD
        }
    }
}

// must input n <= m
fn gcd(n: usize, m: usize) -> usize {
    if n == 0 {
        m
    } else {
        gcd(m%n, n)
    }
}

use std::cmp;
use std::collections::{HashSet, BinaryHeap, VecDeque, HashMap};
use std::iter::FromIterator;
fn main() {
    input! {
        D: usize,
        G: usize,
        problems: [(usize, usize); D],
    }
    let mut min = std::usize::MAX;
    for ind in 0..(1<<D) {
        let mut quo = ind;
        let mut max = (1, 1);
        let mut g = G;
        let mut n = 0;
        for (ind, &(num, comp)) in problems.iter().enumerate() {
            let point = 100 * ind + 100;
            if quo&1 == 1 {
                g = g.saturating_sub(point*num+comp);
                n += num;
            } else {
                if point > max.0 { max = (point, num) }
            }
            quo >>= 1;
        }
        let plus = (g + max.0 - 1) / max.0;
        if plus < max.1 {
            n += plus;
            min = cmp::min(min, n);
        }
    }
    println!("{}", min);
}