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
import java.util.Arrays;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int D = Integer.parseInt(sc.next());
		int G = Integer.parseInt(sc.next());
		int[][] p = new int[D][2];
		for(int i=0; i<D; i++){
			p[i][0] = Integer.parseInt(sc.next());	//問題数
			p[i][1] = Integer.parseInt(sc.next());	//ボーナス点
		}
		
		int ans = Integer.MAX_VALUE;
		
		for(int i=0; i<(1<<D); i++){
			int solve = 0;
			int need = G;
			for(int j=0;j<D;j++){
				if((i>>j & 1)==1){
					solve += p[j][0];
					need -= ((j+1)*100*p[j][0] + p[j][1]);
				}
			}
			
			if(need>0){
				for(int k=D-1; k>=0; k--){
					if(((i>>k) & 1)== 1)continue;
					int count=0;
					while(p[k][0]>count){
						need -= (k+1)*100;
						solve++;
						count++;
						if(need<=0)break;
					}
					if(need>0)solve=Integer.MAX_VALUE;
					break;
				}
			}
			if(solve < ans)ans = solve;
		}		
		System.out.println(ans);
		sc.close();
	}
}
