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)

I=lambda:map(int,input().split())
divup=lambda x,y:(x+y-1)//y
D,G=I()
P={};C={};Score={}
for i in range(D):P[i],C[i]=I();Score[i]=-~i*100*P[i]+C[i]
r=2147483647
for k in range(2**D):
    str_branch=str(bin(k))[2:].rjust(D,'0')
    sub_probs=0;sub_score=0
    for i,b in enumerate(str_branch):
        if b=='1':sub_probs+=P[i];sub_score+=Score[i]
    if sub_score>G:r=min(r,sub_probs);continue
    for i in range(D)[::-1]:
        pi=-~i*100
        if str_branch[i]=='0' and G-sub_score<pi*P[i]:
            r=min(r,sub_probs+divup(G-sub_score,pi));break

r2=2147483647
for k in range(2**D):
    s=''.join((['0']*D+list(str(bin(k))[2:]))[-D:])
    vs=[(P[i]*100*(i+1)+C[i])*int(s[i]) for i in range(D)]
    v=sum(vs)
    tr=sum([P[i]*int(s[i]) for i in range(D)])
    if v>=G:
        r2=min(r,tr)
    else:
        rest_max=D-s[::-1].find('0')-1
        if rest_max<D and v+(P[rest_max]-1)*100*(rest_max+1)>=G and (G-v)//((rest_max+1)*100)<P[rest_max]:
            r=min(r,tr+(G-v)//((rest_max+1)*100))

print(min(r,r2))