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 PyPy3 (2.4.0) (judge verdict: Accepted)

# 入力
D, G = map(int, input().split())
ps = [0] * D
cs = [0] * D

for i in range(D):
    ps[i], cs[i] = map(int, input().split())

# 再帰的に問題数を求める
def dfs(n, d, g):
    """
    d以下の問題でg点以上取るために必要な最小の問題数を返す

    Parameters
    ----------
    n : int
        dより大きいレベルを解いた数
    d : int
        レベル
    g : int
        点数
    m : int
        dレベルを解く場合の問題数
    p : int
        p問解いた場合に得られる点数

    Returns
    -------
    g点取るための問題数
    """
    # d < 1でgに満たない場合
    if d < 1:
        return float('inf')

    # 解くべき問題数(piまたはg点取るのに必要な最小の問題数)
    m = min(-(-g // (100 * d)), ps[d-1])

    # m問解いた時に得られる点数
    p = 100 * m * d + cs[d-1] if m == ps[d-1] else 100 * m * d

    
    if p >= g:  # g点以上取れる
        return min(n+m, dfs(n, d-1, g))
    else:       #g点に満たない
        return min(dfs(n + m, d - 1, g - p), dfs(n, d -1, g))

print(dfs(0, D, G))