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)

from collections import deque
import itertools, math
d,g = list(map(int,input().split()))
pc = [list(map(int,input().split())) for _ in range(d)]
g /= 100

def pattern():
  pat = deque()
  temp = deque()
  for i in sorted(range(d),reverse=True):
    if pc[i][0]*(i+1) >= g:
      pat.append(i)
  
  for i in range(d):
    if i not in pat:
      if pc[i][0]*(i+1) + pc[i][1]//100 >= g:
        temp.append([i,pc[i][0]])
  temp = sorted(temp, reverse=False, key=lambda x: x[1])
  temp = [x[0] for x in temp]
  
  pat.extend(temp)
  for i in sorted(range(d),reverse=True):
    if i not in pat:
      pat.append(i)
  return pat

pat = pattern()
ans = sum([pc[x][0] for x in range(d)])

def dfs(k, point, cnt, pat):
  global ans
  i = pat[k]
  
  cnt += min(pc[i][0], math.ceil((g-point)/(i+1)))
  point += min(pc[i][0], math.ceil((g-point)/(i+1))) * (i+1)
  point += pc[i][1]//100
  #print(k,i,point,cnt)
  if point >= g:  
    ans  = min(ans, cnt)
  else:
    if k < d-1:
      dfs(k+1, point, cnt, pat)
      

dfs(0, 0, 0, pat)
dfs(0,0,0,list(reversed(pat)))

print(ans)