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
#include<iostream>
#include<algorithm>
using namespace std;
static const int MAX=10;
int main(void)
{
	int N,M;
		cin >> N >> M;
	int A[MAX],B[MAX];
	for(int i=0;i<N;i++)
		cin >> A[i] >> B[i];
	
	int ans=(M+99)/100;
	for(int bit=0;bit<(1<<N);bit++){
		int sum=0,cnt=0,m;
		for(int i=0;i<N;i++){
			if(bit&(1<<i)){
				sum+=A[i]*100*(i+1)+B[i];
				cnt+=A[i];
			}
			else
				m=i;
		}
		
		if(sum<M){
			if(M-sum>A[m]*100*(m+1))
				continue;
			cnt+=(M-sum+((m+1)*100)-1)/((m+1)*100);
		}
		ans=min(ans,cnt);
	}
	cout << ans << endl;
	return 0;
}