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.Scanner;

class Main{

	static char[][] map;
	static int H;
	static int W;
	static	int D ;
	static	int G ;
	static	int[] p;
	static	int[] c;
	static int ans = 100000000;

	public static void main(String[] args) {



		Scanner sc = new Scanner(System.in);			//文字の入力
		D = sc.nextInt();
		G = sc.nextInt();
		p = new int[D+1];
		c = new int[D+1];

		for(int i = 1;i < D+1;i++){
			p[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}



		dfs(D-1,p[D],100*D*p[D]+c[D]);
		
		if(G  <= p[D]*(100*D)){				//現在のG-scoreに対し、
			if(G % (100*D) == 0){
				dfs(D-1, (G /(100*D)), G);
			}else{
				dfs(D-1, (G /(100*D))+1, G);

			}

		}
		dfs(D-1,0,0);

		System.out.println(ans);


	}




	static void dfs(int mondai,int sum,int score){			//numは数、mondaiは問題番号,sumは問題をといた数


		if(score >= G){
			
			ans = Math.min(ans, sum);
			return;
		}
		if(mondai < 1){
			return;
		}


		dfs( mondai-1, sum+p[mondai], score+100*mondai*p[mondai]+c[mondai]);

		if((G-score)  <= p[mondai] *(100*mondai) ){				//現在のG-scoreに対し、

			if((G-score) %(100*mondai)== 0){
				dfs(mondai-1, sum+ ((G-score) /(100*mondai)), G);
			}else{
				dfs(mondai-1, sum+ ((G-score) /(100*mondai))+1, G);

			}

		}

		dfs( mondai-1, sum, score);
		return;
	}


}



