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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
list($N, $G) = ints();
for($i = 0; $i < $N; $i++){
    list($C[], $B[]) = ints();
}
$min = 10**10;
for($i = 0; $i < 2**$N; $i++){
    $c = 0;
    $score = 0;
    $bits = sprintf("%0{$N}b", $i);
    for($j = 0; $j < $N; $j++){
        if($bits[$j]){
            $c+=$C[$j];
            $score+=$C[$j]*($j+1)*100+$B[$j];
        }
    }
    for($j = $N-1; $j >= 0; $j--){
        if($bits[$j] == 0){
            for($k = 0; $k < $C[$j]-1; $k++){
                if($score < $G){
                    $c++;
                    $score+=($j+1)*100;
                }
            }
        }
    }
    if($score>=$G){
        chmin($min, $c);
    }
}
echo $min,"\n";

function str(){return trim(fgets(STDIN));}
function ints(){return array_map('intval',explode(' ',trim(fgets(STDIN))));}
function int(){return intval(trim(fgets(STDIN)));}
function chmax(&$a,$b){if($a<$b){$a=$b;return 1;}return 0;}
function chmin(&$a,$b){if($a>$b){$a=$b;return 1;}return 0;}
function popcount($x){$c=0;while($x){$x&=$x-1;++$c;}return$c;}
function o(...$val){
    if(count($val)==1)$val=array_shift($val);
    echo debug_backtrace()[0]['line'].")";
    if(is_array($val)){
        if(count($val) == 0)echo "empty array";
        elseif(!is_array(current($val)))echo "array: ",implode(" ", addIndex($val)),"\n";
        else{
            echo "array:array\n";
            if(isCleanArray($val))foreach($val as $row)echo implode(" ", addIndex($row)),"\n";
            else foreach($val as $i => $row)echo "[".$i."] ".implode(" ", addIndex($row)),"\n";
        }
    }else echo $val."\n";
}
function addIndex($val){if(!isCleanArray($val)){$val = array_map(function($k, $v){return $k.":".$v;}, array_keys($val), $val);}return $val;}
function isCleanArray($array){$clean=true;$i = 0;foreach($array as $k=>$v){if($k != $i++)$clean = false;}return $clean;}
// 座圧対象の配列 -> [圧縮された配列,復元用のMap,圧縮用のMap]
function atsu($array){$a = array_flip($array);$fuku=array_flip($a);sort($fuku);$atsu = array_flip($fuku);foreach($array as $i=>$val)$array[$i]=$atsu[$val];return [$array, $fuku, $atsu];}
function gcdAll($array){$gcd=$array[0];for($i=1,$I=count($array);$i<$I;++$i){$gcd=gcd($gcd,$array[$i]);}return $gcd;}
function gcd($m, $n){if(!$n)return $m;return gcd($n, $m % $n);}
function lcmAll($array){$lcm=$array[0];for($i=1,$I=count($array);$i<$I;++$i){$lcm=lcm($lcm,$array[$i]);}return $lcm;}
function lcm($a, $b) {return $a / gcd($a, $b) * $b;}

function dfs($current, $from = -1, $d = 0){
  global $G, $distance;
  $distance[$current] = $d;
  foreach($G[$current] as $to){
    if($from != $to){
      dfs($to, $current, $d + 1);
    }
  }
}

function loadTree($N = false){
  if($N === false)$N = $GLOBALS['N'];
  return loadGraph($N, $N-1);
}
function loadGraph($N = false, $M = false, $both = true){
  if($N === false)$N = $GLOBALS['N'];
  if($M === false)$N = $GLOBALS['M'];
  $G = array_fill(1, $N, []);
  for($i = 0; $i < $M; $i++){
    $values = ints();
    if(count($values) == 2){
      list($a, $b) = $values;
      $G[$a][] = $b;
      if($both)$G[$b][] = $a;
    }else{
      list($a, $b, $d) = $values;
      $G[$a][] = [$b, $d];
      if($both)$G[$b][] = [$a, $d];
    }
  }
  return $G;
}