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 Python (3.4.3) (judge verdict: Wrong Answer)

import math

d, g = [int(v) for v in input().split()]
#print(d, g)
pc_s = [[int(v) for v in input().split()] for _ in range(d)]
for i in range(d):
  pc = pc_s[i]
  pc_s[i].append((i + 1) * 100)
  pc_s[i].append( (pc[0] * pc[2] + pc[1]) / pc[0])
#print(pc_s)

#1問あたり得点順
pc_s = pc_s[::-1]
#1問あたりの得点＋ボーナス順
a_s = pc_s[:]
a_s.sort(key = lambda x:x[3], reverse=True)
#問題数順
cnt_s = pc_s[:]
cnt_s.sort(key = lambda x:x[0])

#print(pc_s)
#print(a_s)
solved = 0
score = 0
while score < g:
  cand_solved1 = 0
  cand_solved2 = 0
  
  #ボーナスありでクリア
  for pc in cnt_s:
    if pc[0] > 0:
      if pc[0] * pc[2] + pc[1] + score >= g:
        cand_solved1 = pc[0]
        #print("cand1={}".format(cand_solved1))
        break
  
  #ボーナスなしでクリア
  for pc in pc_s:
    #print("pc0[{}]*pc2[{}]+score[{}], g[{}]".format(pc[0], pc[2], score, g))
    if pc[0] > 0:
      if pc[0] * pc[2] + score >= g:
        cand_solved2 = math.ceil((g - score) / pc[2])
        #print("cand2={}".format(cand_solved2))
        break

  #クリアした場合はどちらを採用するか検討
  if cand_solved1 > 0 or cand_solved2 > 0:
    #print(cand_solved1, cand_solved2)
    if cand_solved1 == 0:
      num = cand_solved2
    elif cand_solved2 == 0:
      num = cand_solved1
    else:
      num = min(cand_solved1, cand_solved2)     
    score += g
    solved += num

  #高効率な問題をとく
  if score < g:
    #print("a_s", a_s)
    for pc in a_s:
      if pc[0] > 0:
        #print("最高効率:", pc)
        score += pc[0] * pc[2] + pc[1]
        solved += pc[0]
        pc[0] = 0
        break
  
print(solved)