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;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int d = sc.nextInt();//nanido
		int g = sc.nextInt();//mokuhyou
		int p[] = new int[d];//point
		int c[] = new int[d];//comp point
		int cnt = 0;//mondaisu
		int cntmin = 0;
		int sum = 0;
		for (int i = 0; i < d; i++) {
			p[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}
		for (int bit = 0; bit < (1 << d); bit++) {
			for (int i = 0; i < d; i++) {
				if ((bit & (1 << i)) != 0) {
					sum += p[i] * 100 * (i + 1) + c[i];
					cnt += p[i];
				}
			}
			for (int i = d - 1; i >= 0; i--) {
				if ((bit & (1 << i)) == 0) {
					for (int j = 0; j < p[i] - 1; j++) {
						if (sum >= g) {
							break;
						} else {
							sum += 100 * (i + 1);
							cnt++;
						}
					}
				}
			}
			if (sum >= g && (cntmin == 0 || cnt < cntmin)) {
				cntmin = cnt;
			}
			//System.out.println(cnt);
			cnt = 0;
			sum = 0;
		}
		System.out.println(cntmin);
	}
}
