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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int D=sc.nextInt();
        int g=sc.nextInt();
        int[] ar=new int[D];
        int[] ar1=new int[D];
        for(int i=0;i<D;i++){
            ar[i]=sc.nextInt();
            ar1[i]=sc.nextInt();
        }
        int p=0,ans=Integer.MAX_VALUE;
        int G=0;
        for(int i=D-1;i>=0;i--){
            //System.out.println("i="+i);
            G=g;
            int j=i;
            while(G>0){
                if(j<0){
                    break;
                }
                //System.out.println("A");
                int total=ar[j]*100*(j+1)+ar1[j];
                //System.out.println("j="+j);
                if(total>=G){
                    //System.out.println("B");
                    total-=ar1[j];
                    if(total>=G){
                        //System.out.println("C");
                        p+=ar[j];
                        //System.out.println("total="+total+" G="+G);
                        //System.out.println("p="+p);
                        while(total>G){
                            total-=100*(j+1);
                            p--;
                            //System.out.println("p="+p);
                        }
                        G=0;
                    }
                    else{
                        //System.out.println("D");
                        G=0;
                        p+=ar[j];
                        //System.out.println("p="+p);
                    }
                }
                else{
                    //System.out.println("E");
                    //System.out.println("total="+total);
                    G-=total;
                    p+=ar[j];
                    j--;
                    //System.out.println("p="+p);
                    //System.out.println("G="+G);
                }
                if(G<=0){
                    ans=Math.min(ans,p);
                    //System.out.println("ans="+ans);
                    p=0;
                }
            }
        }
        System.out.println(ans);
    }
}