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: Accepted)

def min(a,b); a < b ? a : b; end
def max(a,b); a > b ? a : b; end

D, G = gets.split.map(&:to_i)
P = (1 .. D).map do |w|
  c,b = gets.split.map(&:to_i)
  [w,c,b]
end

ans = (1 << D).times.inject(1000_000) do |min_solve, complete|
  unused, used = D.times.partition{|i| complete[i].zero? }

  solved_offset = used.inject(0){|s,i| s + P[i][1]}
  next min_solve if solved_offset >= min_solve
  
  point = used.inject(0){|s, i| s + 100 * P[i][0] * P[i][1] + P[i][2] }
  next min(min_solve, solved_offset) if point >= G
  
  rest = unused.reverse_each.inject([0, G - point]) do |s, i|
    solved_count, rest = s
    w,c,b = P[i]
    r = min((rest + 100 * w - 1) / (100 * w), c - 1)
    if rest - 100 * w * r <= 0
      break [solved_count + r, 0]
    else
      [solved_count + r, rest - 100 * w * r]
    end
  end
  
  if rest[1].zero?
    min(min_solve, solved_offset + rest[0]) 
  else
    min_solve
  end
  
end

puts ans