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 PHP (5.6.30) (judge verdict: Wrong Answer)

<?php
//error_reporting(0);
//$S = trim(fgets(STDIN));
list($D,$G) = explode(" ",trim(fgets(STDIN)));
//$a = explode(" ",trim(fgets(STDIN)));
for($i=0;$i<$D;$i++){
  list($P[],$C[])= explode(" ",trim(fgets(STDIN)));
}
$tot = 0;
$res = 99999;
//どれかを解くだけで満たす場合
for($i=0;$i<$D;$i++) {
  //printf("%d %d %d\n",$i+1,$P[$i],$G/(($i+1)*100));
  if (($G / (($i+1)*100))<$P[$i]) {
    $res = min($res,$G / (($i+1)*100));
  }
}
$comp=array();
//どれかをコンプリートすると満たす場合
for($i=0;$i<$D;$i++) {
  $sum = ($i+1)*100*$P[$i]+$C[$i];
  //printf("%d %d %d\n",$i+1,$P[$i],$sum);
  if ($sum>=$G) {
    $res = min($res,$P[$i]);
  }
  $comp[$i]=$sum/$P[$i];
}
arsort($comp);
//コンプリート点数/回数が高いものからやってみる
$tot = 0;
$cnt = 0;
foreach($comp as $i => $v) {
  for($j=0;$j<$P[$i];$j++) {
    $tot += (($i+1)*100);
    $cnt++;
    if ($tot >= $G) {
      break;
    }
  }
  $tot+=$C[$i];
  if ($tot >= $G) {
    break;
  }
}
printf("%d\n",min($cnt,$res));