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.*;
public class Main {
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int D = sc.nextInt();
		int G = sc.nextInt();
		int p[] = new int[D];
		int c[] = new int[D];
		for(int i = 0; i < D; i++) {
			p[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}
		
		int ans = Integer.MAX_VALUE;
		out : for (int i = 0; i < (Math.pow(2, D)); i++) {
			int temp_ans = 0;
			boolean flag[] = new boolean[D];
			Arrays.fill(flag, false);
			int sum = 0;
			for (int j = 0; j < D; j++) {
				if ((1&i>>j) == 1) {
					sum += p[j] * (j + 1) * 100;
					sum += c[j];
					flag[j] = true;
					temp_ans += p[j];
				}
			}
			if(sum >= G) {
				ans = Math.min(ans, temp_ans);
				continue out;
			}
			in : for(int j = 0; j < D; j++) {
				if(!flag[D - 1 - j]) {
					for(int k = 0; k < p[D - 1 - j]; k++) {
						sum += 100 * (D - j);
						temp_ans++;
						if(sum >= G) {
							ans = Math.min(ans, temp_ans);
							break in;
						}
					}
					//break in;
				}
			}
		}
		System.out.println(ans);
	}
}