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.Arrays;
import java.util.Scanner;

public class Main {
    void run() {
        Scanner sc = new Scanner(System.in);
        int d = sc.nextInt(), g = sc.nextInt();
        int[] ps = new int[d], cs = new int[d];

        for (int i = 0; i < d; i++) {
            ps[i] = sc.nextInt();
            cs[i] = sc.nextInt();
        }

        int L = 100 * 10;
        long[] dp = new long[L + 1];

        Arrays.fill(dp, -1);
        dp[0] = 0;

        for (int i = 0; i < d; i++) {
            for (int j = L; j >= 0; j--) {
                if (dp[j] >= 0) {
                    for (int k = 1; k < ps[i]; k++) {
                        dp[j + k] = Math.max(dp[j + k], dp[j] + k * (i + 1) * 100);
                    }
                    dp[j + ps[i]] = Math.max(dp[j + ps[i]], dp[j] + ps[i] * (i + 1) * 100 + cs[i]);
                }
            }
        }
//        debug(dp);
        for (int i = 1; i <= L; i++) {
            if (dp[i] >= g) {
                System.out.println(i);
                return;
            }
        }
    }

    void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }

}
