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)

lines = readlines.map(&:strip)

n, d = lines.shift.split.map(&:to_i)

problems = lines.map{|line|line.split.map(&:to_i)}

p [n, d, problems].inspect

# I need calculate rest count with biggest problems after current point
def count_problems_in_rest(complete_solving_arr, problems, goal)
  diff = goal - points_with_solved(complete_solving_arr, problems)
  return 0 if diff<0

  last_index = complete_solving_arr.zip((1..complete_solving_arr.count).to_a).keep_if{|e|e[0]==0}.last&.last.to_i
  max_points_in_rest = last_index*100
  return 0 if last_index == 0

  require_count = (diff.to_f/max_points_in_rest.to_f).ceil
  return Float::INFINITY if require_count > problems[last_index-1][0]
  require_count
end

# calculate point with current solving array
def points_with_solved(complete_solving_arr, problems)
  complete_solving_arr.zip(problems).map.with_index {|solve_problem, index|
    solve= solve_problem[0]
    problem = solve_problem[1]
    if(solve==1)
      100*(index+1)*problem[0]+problem[1]
    else
      0
    end
  }.inject(:+)
end

def count_with_solved(solvings, problems)
  solvings.zip(problems).keep_if{|e|e[0]==1}.map{|e|e[1][0]}.inject(:+).to_i
end

def recursive_count(solvings, problems, goal)
  # end condition
  if solvings.length==problems.length
    current_points = points_with_solved(solvings, problems)
    count = count_with_solved(solvings, problems)
    additional_count = count_problems_in_rest(solvings, problems, goal)

#p [count+additional_count, solvings]

    return count+additional_count
  end

  # recursive condition
  return [
    recursive_count(solvings.dup.push(0), problems, goal), 
    recursive_count(solvings.dup.push(1), problems, goal)
  ].min
end


#puts points_with_solved([0,1], problems)
#puts count_problems_in_rest([0,1], problems, 7000).inspect

p recursive_count([], problems, d)