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
    public class Main {
        private static java.util.Scanner scanner = new java.util.Scanner(System.in);
     
        public static void main(String[] args) {
            int d = scanner.nextInt(), g = scanner.nextInt();
            int[] p = new int[d], c = new int[d];
            for (int i = 0; i < d; i++) {
                p[i] = scanner.nextInt();
                c[i] = scanner.nextInt();
            }
     
            int ans = 10000;
            for (int i = 0; i < 1 << d; i++) {
                int sum = 0, count = 0, zero = 0;
                for (int j = 0; j < d; j++) {
                    if ((1 << j & i) != 0) {
                        sum += (j + 1) * 100 * p[j] + c[j];
                        count += p[j];
                    } else {
                        zero = j;
                    }
                }
                if (sum < g) {
                    int t = (int)Math.ceil((g - sum) / ((zero + 1) * 100F));
                    if (p[zero] < t)
                        continue;
                    count += Math.ceil((g - sum) / ((zero + 1) * 100F));
                }
                ans = Math.min(ans, count);
            }
            System.out.println(ans);
        }
    }