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 {
    private static final int INF = 10000;
    private static int[][] memo;

    private static int solve(int d, int g, int[] p, int[] c, int used) {
        if(memo[g][used] != 0) return memo[g][used];

        int res = INF;
        for(int i = 0; i < d; i++) {
            int bit = (int) Math.pow(2, i);
            if((used / bit) % 2 == 1) continue;

            if((i + 1) * (p[i] - 1) >= g) {
                res = Math.min(res, (int) Math.ceil(1.0 * g / (i + 1)));
            }else if ((i + 1) * p[i] + c[i] >= g) {
                res = Math.min(res, p[i]);
            }else {
                res = Math.min(res, p[i] + solve(
                    d, g - ((i + 1) * p[i] + c[i]), p, c, used + bit
                ));
            }
        }
        memo[g][used] = res;
        return res;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int d = sc.nextInt();
        int g = sc.nextInt() / 100;

        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() / 100;
        }

        memo = new int[105501][1024];
        System.out.println(solve(d, g, p, c, 0));

        sc.close();
    }
}