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 = dfs(0, new boolean[d], p, c, g, 0);
        System.out.println(ans);

    }

    public static int dfs(int money, boolean[] use, int[] p, int[] c, int g, int u) {
        int min = 99999999;
        for (int id = 0; id < p.length; id++) {
            if (use[id]) {
                continue;
            }
            // 終了する時
            int tmp = g - money;
            int mone = (id + 1) * 100;
            int need = tmp / mone + (tmp % mone == 0 ? 0 : 1);
            if (need > p[id] && mone * p[id] + c[id] >= tmp) {
                min = Math.min(min, u + p[id]);
            } else if (need <= p[id]) {
                min = Math.min(min, u + need);
            } else {
                use[id] = true;
                min = Math.min(dfs(money + p[id] * mone + c[id], use, p, c, g, u + p[id]), min);
                use[id] = false;
            }
        }

        return min;
    }

}
