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: Time Limit Exceeded)

#!/usr/bin/env python3
import sys, math, fractions, itertools


def solve(D: int, G: int, p: "List[int]", c: "List[int]"):
    mn = 1e5
    for i in range(1 << D):
        score = 0
        count = 0
        for j in range(D):
            if i >> j & 1 == 1:
                score += p[j] * 100 * (j+1) + c[j]
                count += p[j]
        while(score < G):
            flag = True
            for k in range(D):
                j = D - k - 1
                if i >> j & 1 == 0:
                    if score + 100*(j+1)*(p[j]-1) >= G:
                        count += math.ceil((G - score) / (100*(j+1)))
                        flag = False
                        break
                    else:
                        score += 100*(j+1)*(p[j]-1)
                        count += p[j] - 1
            if not flag:
                break
        mn = min(mn, count)
    print(mn)
    return


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    D = int(next(tokens))  # type: int
    G = int(next(tokens))  # type: int
    p = [int()] * (D)  # type: "List[int]" 
    c = [int()] * (D)  # type: "List[int]" 
    for i in range(D):
        p[i] = int(next(tokens))
        c[i] = int(next(tokens))
    solve(D, G, p, c)

if __name__ == '__main__':
    main()