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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# AtCoder Beginner Contest 104
# C - All Green
# https://atcoder.jp/contests/abc104/tasks/abc104_c
ENV[Z='RUBY_THREAD_VM_STACK_SIZE']||exec({Z=>'100000000'},'ruby',$0)
INF = 10**19
(D, G) = gets.split.map(&:to_i)
point = 0
@scores = readlines.map{|l| l.chomp.split.map(&:to_i)}
@scores = @scores.map{|s| [point += 100] + s}.reverse

def dfs(i, sum)
  #pp [i, sum]
  return INF if i == @scores.length
  return 0 if G <= sum
  if (G - sum) / @scores[i][0] <= @scores[i][1]
    return (G - sum) / @scores[i][0]

  elsif @scores[i][0] * @scores[i][1] + @scores[i][2] > (G - sum)
    return @scores[i][1]

  elsif (G - sum) > @scores[i,@scores.length].reduce(0){|sum,s| sum+s[0]*s[1]+s[2]}
    return INF
  else
    counts = []
    # 全部使う
    counts << @scores[i][1] + dfs(i+1, sum+@scores[i][0]*@scores[i][1]+@scores[i][2])

    # 使わない
    counts << dfs(i+1, sum)

    # max - 1 使う
    (1...@scores[i][1]).each do |n|
      cnt = @scores[i][1]-n + dfs(i+1, sum+@scores[i][0]*(@scores[i][1]-n))
      break if cnt > INF
      #break if counts.last < cnt
      counts << cnt
    end

    #pp ["count#{i}"] + counts
    return counts.min
  end
end

puts dfs(0, 0)