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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
D,G=map(int,raw_input().split())

A=[map(int,raw_input().split()) for i in range(D)]

N=1000
dp=[ [ float("-inf") for i in range(N+100)]  for j in range(D) ] 

p=A[0][0]
c=A[0][1]


for i in range(1,p+1):
	if i==p: dp[0][i]=i*100+c
	else: dp[0][i]=i*100

for i in range(D):
	dp[i][0]=0



for k in range(1,D):
	p=A[k][0]    # number of problems per line
	c=A[k][1]    # bonus

	for i in range(N):		         #dp[k][i]
		for j in range(1,p+1):    #number of problems per line
			dp[k][i+j]=max( dp[k][i+j], dp[k-1][i]+j*(k+1)*100, dp[k-1][i+j] )
		dp[k][i+j]=max( dp[k][i+j], dp[k-1][i]+j*(k+1)*100+c,dp[k-1][i+j] )



for idx,i in enumerate(dp[D-1]):
	if i>=G:
		print idx
		quit()
		
