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>
using namespace std;

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()

typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;



int main(){
	int d,g,p[11],c[11];
	cin>>d>>g;
	for (int i = 0; i < d; i++){
		cin>>p[i]>>c[i];
	}

	int ans=1e+9;
	for (int i = 0; i < (1<<d); i++){
		int s=0,num=0,res=-1;
		for (int j = 0; j < d; j++){
			if(i&(1<<j)){
				s+=100*(j+1)*p[j]+c[j];
				num+=p[j];
			}else{
				res=j;
			}
		}
		if(s<g){
			int s1=100*(res+1);
			int need=(g-s+s1-1)/s1;
			if(need>=p[res])continue;
			num+=need;
		}
		ans=min(ans,num);
	}

	cout<<ans<<endl;   
	return 0;
}