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<bits/stdc++.h>
#define f(i,n) for(int i=0;i<(n);i++)
using namespace std;
signed main(){
	int a,b,c[10],d[10],ans=INT_MAX;
	cin>>a>>b;
	f(i,a) cin>>c[i]>>d[i];
	f(i,pow(2,a)){
		int x=b,y=0,z=i;
		bool vz[10]={true,true,true,true,true,true,true,true,true,true};
		f(j,a){
			if(z%2){
				x-=(c[j]*(j+1)*100+d[j]);
				y+=c[j];
				vz[j]=false;
			}
			z/=2;
		}
		int vx=a-1;
		int vy=c[vx]-1;
		while(x>0&&vx>=0){
			if(vz[vx]&&vy){
				x-=vx*100+100;
				y++;
				vy--;
			}
			else{
				vx--;
				if(vx>=0) vy=c[vx]-1;
			}
		}
		if(x<=0) ans=min(ans,y);
	}
	cout<<ans<<endl;
	return 0;
}
