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
import copy
D, G = map(int, input().split())
pc = [list(map(int, input().split())) for _ in range(D)]
aveP = [[(100*(i+1)*pc[i][0]+pc[i][1])/pc[i][0], i, pc[i][0], (100*(i+1)*pc[i][0]+pc[i][1])] for i in range(D)]
aveP2 = aveP
ans = 0
for i in range(D):
    if G <= 100*(i+1)*pc[i][0]:
        aveP2[i][0] = 100*(i+1)
aveP2.sort()
aveP3 = copy.deepcopy(aveP2)
for i in range(D):
    c = aveP3[i][1]
    aveP3[i][1] = aveP3[i][0]
    aveP3[i][0] = c
aveP3.sort()
p = 0
for i in range(1,D+1):
    for l in range(1,len(aveP3)+1):
        if G <= 100 * (aveP3[-l][0] + 1) * aveP3[-l][2] and aveP2[-i][2] > math.ceil(G / (100 * (aveP3[-l][0] + 1))) :
            ans += math.ceil(G / (100 * (aveP3[-l][0] + 1)))
            p = 1
            break
    if p == 1:
        break
    G -= aveP2[-i][3]
    ans += aveP2[-i][2]
    for l in range(len(aveP3)):
        if aveP3[l][0] == aveP2[-i][1] and aveP3[l][1] == aveP2[-i][0] and aveP3[l][2] == aveP2[-i][2] and aveP3[l][3] == aveP2[-i][3]:
            del aveP3[l]
            break
    if G <= 0:
        break
print(ans)