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: Time Limit Exceeded)

use std::fmt;
use std::io::stdin;
use std::io::BufRead;
use std::str;
use std::str::FromStr;
use std::vec::Vec;
use std::cmp;

fn main() {
    let stdin = stdin();
    let mut scan = InputScanner::new(stdin.lock(),256);

	let d:usize = scan.next();
	let g:usize = scan.next();

	let pc:Vec<(usize,usize)> = (0..d).map(|_| (scan.next(),scan.next())).collect::<Vec<(usize,usize)>>();

	let answer = search(d,g as i64,&pc,0,0);

	println!("{}",answer);
}
fn search(d:usize,g:i64,pc:&[(usize,usize)],i:usize,count:usize) -> usize {
	let mut min:usize = std::usize::MAX;

	let g = g as i64;

	if i == d {
		if g <= 0 {
			println!("{}",count);
			return count;
		} else {
			return std::usize::MAX;
		}
	}

	if g - (pc[i].0 * (i+1) * 100 + pc[i].1) as i64 <= 0 {
		min = count + pc[i].0;
	} else {
		let n = pc[i].0;
		min = cmp::min(min,search(d,g-(n*(i+1)*100 + pc[i].1) as i64,pc,i+1,count+n));
	}

	for n in (0..pc[i].0).rev() {
		if g - (n * (i+1) * 100) as i64 <= 0 {
			min = count + n;
			continue;
		}
		min = cmp::min(min,search(d,g-(n*(i+1)*100) as i64,pc,i+1,count+n));
	}

	min
}
struct InputScanner<R: BufRead> {
	reader: R,
	buf: Vec<u8>, // Should never be empty
	pos: usize,   // Should never be out of bounds as long as the input ends with '\n'
}
impl<R: BufRead> InputScanner<R> {
	fn new(reader: R, capacity: usize) -> Self {
		InputScanner {
			reader: reader,
			buf: Vec::with_capacity(capacity),
			pos: 0,
		}
	}

	#[inline]
	fn next<T: FromStr>(&mut self) -> T
	where
		T::Err: fmt::Debug,
	{
		if self.buf.is_empty() {
			self._read_next_line();
		}
		let mut start = None;
		loop {
			match (self.buf[self.pos], start.is_some()) {
				(b' ', true) | (b'\n', true) | (b'\r', true) => break,
				(_, true) | (b' ', false) => self.pos += 1,
				(b'\n', false) | (b'\r', false) => self._read_next_line(),
				(_, false) => start = Some(self.pos),
			}
		}
		let target = &self.buf[start.unwrap()..self.pos];
		unsafe { str::from_utf8_unchecked(target) }.parse().unwrap()
	}

	#[inline]
	fn _read_next_line(&mut self) {
		self.pos = 0;
		self.buf.clear();
		if self.reader.read_until(b'\n', &mut self.buf).unwrap() == 0 {
			panic!("Reached EOF");
		}
	}
}