Problem:
Score : 300 points

Problem Statement

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).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.text.*;

public class Main {

    Scanner sc = new Scanner(System.in);
    int d = sc.nextInt();
    int g = sc.nextInt();
    int[][] arr = new int[d][2];
    int res = (int)1e5;

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

    void run() {

        for (int i=0; i<d; i++) {
            arr[i][0] = sc.nextInt();
            arr[i][1] = sc.nextInt();
        }

        dfs("");

        System.out.println(res);

    }

    void dfs(String s) {
        if (s.length()==d) {
//            System.out.println(s);
            int sum = 0;
            int cnt = 0;
            for (int i=0; i<d; i++) {
                if (s.charAt(i)=='1') {
                    sum += arr[i][0] * 100 * (i+1) + arr[i][1];
                    cnt += arr[i][0];
                }
            }
            sum = g - sum;
//            System.out.println(sum);
            if (sum<=0) {
//                System.out.println(cnt);
                res = Math.min(res, cnt);
            } else {
                for (int i=d; 1<=i; i--) {
                    if (s.charAt(i-1)=='0') {
                        int max = arr[i-1][0]-1;
                        int need = sum%(100*i)==0 ? sum/(100*i) : (sum/(100*i))+1;
                        if (need<=max) {
                            cnt += need;
                            sum -= (100*i)*need;
                            if (sum<=0) break;
                        } else {
                            sum -= (100*i)*max;
                            cnt += max;
                            if (sum<=0) break;
                        }
                    }
                }
//                System.out.println(cnt);
                if (sum<=0) res = Math.min(res, cnt);
            }

        } else {
            for (int i=0; i<2; i++) {
                dfs(s + i);
            }
        }
    }

//    void dfs(String s) {
//        if (split(s, 'T')==d-1) {
//            System.out.println(s);
//            String[] ts = s.split("T", 0);
//            int[] a = new int[d];
//            int cnt = 0;
//            for (int i=0; i<d; i++) a[i] = Integer.parseInt(ts[i]);
//            int sum = 0;
//            for (int i=0; i<d; i++) {
//                cnt += a[i];
//                sum += a[i] * 100 * (i+1);
//                if (a[i]==arr[i][0]) sum += arr[i][1];
//            }
//            if (g<=sum) res = Math.min(res, cnt);
//        }
//        else {
//            for (int i=0; i<d; i++) {
//                for (int j=0; j<=arr[i][0]; j++) {
//                    dfs(s + j + "T");
//                }
//            }
//        }
//    }
//
//    int split(String s, char c) {
//        return s.split(c + "").length - 1;
//    }


}