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>
#include <vector>
#include <math.h>
using namespace std;
typedef long long int ll;

int main(){
	int d,g; cin >> d >> g;
	vector<int> p(d),c(d);
	for(int i=0;i<d;i++){
		cin >> p[i] >> c[i];
	}
	int ans=1e9;
	for(int i=0;i<(1<<d);i++){
		int last=-1;
		int sum=0; int ad=0;
		for(int j=0;j<d;j++){
			if((i>>j)&1){
				sum+=(c[j]+(p[j]*(1+j)*100));
				ad+=p[j];
			}
			else last=j;
		}
		if(sum>=g){
			ans=min(ans,ad);
		}
		else if(p[last]*(last+1)*100+sum<g)continue;
		else{
			sum-=g;
			sum*=(-1);
			ad+=(sum+(last+1)*100-1)/((last+1)*100);
			ans=min(ans,ad);
		}
	}
	cout << ans << endl;
}