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 Perl (v5.18.2) (judge verdict: Time Limit Exceeded)

#! /usr/bin/perl

use strict;
# use Clone qw(clone);
use List::Util qw(min);
use POSIX qw(ceil);

my @lines = <STDIN>;
@lines = map { $_ =~ s/^\s*(.*)\s*$/$1/; $_ } @lines;

my ($D, $G) = split(/ /, $lines[0]);
$G /= 100;
my @pc = map { my @tmp = split(/ /, $_);
               $tmp[1] /= 100;
               \@tmp } @lines[1..$#lines];

sub solve {
    my $G = $_[0];
    my $pc = $_[1];

    if ($G <= 0) {
        return 0;
    }

    my $min;
    
    for my $i (0..$#$pc) {
        if ($pc->[$i][0] != 0) {
            my $pc2 = [map { [@$_] } @$pc]; # clone($pc);
            $pc2->[$i][0] = 0;
            pop(@$pc2) while (@$pc2 && $pc2->[$#$pc2][0] == 0);

            my $ret = $pc->[$i][0] + solve( $G - ($i+1)*$pc->[$i][0] - $pc->[$i][1], $pc2);
            $min = defined($min)
                ? min($min, $ret) : $ret;
        }
    }

    if ($G / ($#$pc+1) <= $pc->[$#$pc][0]) {
        $min = min($min, ceil($G / ($#$pc+1)));
    }

    return $min;
}

print(solve($G, \@pc), "\n");