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 <utility>
#include <algorithm>
#include <vector>
using namespace std;
int D,G,P[11],C[11];

int main(){
	cin >> D >> G;
	vector<pair<int,int>> v;
	for(int i=0;i<D;i++){
		cin >> P[i] >> C[i];
	}	
	int ans = 1e9;
	for(int i=0;i<(1<<D);i++){
		int preans = 0,id = -1,score = 0;
		for(int j=0;j<D;j++){
			if(i & (1<<j)){
				preans += P[j];
				score += C[j]+P[j]*100*(j+1);
			}else{
				id = j;
			}
		}
		if(id>=0){
			int count = 0;
			while(count<P[id]-1 && score<G){
				score += 100*(id+1);
				count++;
				preans++;
			}
		}
		if(score>=G) ans = min(preans,ans);
	//	cout << id << " " << score << endl;
	}
	cout << ans << endl;
}