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: Wrong Answer)

import java.util.*;
class Main
{
    public static void main (String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int d = sc.nextInt();
        long G = sc.nextLong() / 100;
        int[] p = new int[d], g = new int[d], max = new int[d];
        int k = (int)Math.pow(2, d);
        for(int i = 0; i < d; i++){
            p[i] = sc.nextInt();
            g[i] = sc.nextInt() / 100;
            max[i] = (i + 1) * p[i] + g[i];
        }
        int ans = Integer.MAX_VALUE;
        for(int i = 0; i < k; i++){
            long gg = G;
            String kk = to2(i, d);
            int tmp = 0;
            int min = -1;
            for(int j = 0; j < d; j++){
                if(kk.charAt(j) == '1'){
                    gg -= max[j];
                    tmp += p[j];
                }else{
                    min = j;
                }
            }
            if(min >= 0){
                if(0 <= gg && gg <= (min + 1) * p[min])
                    tmp += Math.min((gg + min)/ (min + 1), p[min] - 1);
                else tmp = Integer.MAX_VALUE;
            }
           else if(min == -1){
               if(gg > 0) tmp = Integer.MAX_VALUE;
           }
            ans = Math.min(ans, tmp);
        }
        System.out.println(ans);
    }
    static String to2 (int m, int d){
        String ans = "";
        for(int i = 0; i < d; i++){
            if(m % 2 == 0){
                m /= 2;
                ans = "0" + ans;
            }
            else{
                m--;
                m /= 2;
                ans = "1" + ans;
            }
        }
        return ans;
    }
}