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.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int D = Integer.parseInt(sc.next());
    int G = Integer.parseInt(sc.next());
    int[] p = new int[D];
    int[] c = new int[D];

    for (int i=0; i < D; i++) {
      p[i] = Integer.parseInt(sc.next());
      c[i] = Integer.parseInt(sc.next());
    }
    int min = 100;
    for(int i=0; i<Math.pow(2,D); i++){
      int sum = 0; // 点数の合計
      int num = 0;//  解いた問題の合計
      int Most = -1; 
      for(int j=0; j<D; j++){
        System.out.println(i + ":i, " +j + ":j");
        if((i>>j&1)==1){
          num = num + p[j];
          sum = sum + p[j]*(j+1)*100 + c[j];
          System.out.println(num + ":num, " + sum + ":sum ");
        }else if((i>>j&1)==0){
          Most = j + 1;
        }
      int tmp = 0;
      if(sum>=G){
        tmp = num;
      }else if (Most == -1){
        break;
      }else{
        tmp = (G - sum)/(Most*100) + num; 
      }
      if(tmp < min){
        min = tmp;
      }
      System.out.println(min + " :ans");
    }
  }
  System.out.println(min);
  return;
}
}