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.awt.Point;
import java.util.ArrayDeque;
import java.util.Scanner;
import java.util.*;
import java.util.Arrays;
import java.util.ArrayList;
import java.util.Set;
import java.util.HashSet;
import java.util.Collection;


 
public class Main{
	
        static final Scanner sc=new Scanner(System.in);
        
        
        
        
        
        
        public static void main(String[]args){
                int D=sc.nextInt();
                int G=sc.nextInt();
                int[]p=new int[D];
                int[]c=new int[D];
                for(int i=0;i<D;i++){
                        p[i]=sc.nextInt();
                        c[i]=sc.nextInt();
                }
                int[]point=new int[D];
                for(int i=0;i<D;i++){
                        point[i]=100*(i+1)*p[i]+c[i];
                }
                int sum=0;
                int ima=D-1;
                int nokori=p[ima];
                for(;;){
                        
                        if(G<=0){
                                break;
                        }
                        
                        
                        for(int i=0;i<D;i++){
                                if(point[i]>=G){
                                if(p[i]<=p[ima]){
                                        if(point[i]>100*p[i]*(ima+1)){
                                                for(int j=0;j<=p[i];j++){
                                                        if(100*(ima+1)*j>=G){
                                                                break;
                                                        }
                                                        if(j==p[i]){
                                                                if(p[i]!=0){
                                                                ima=i;}
                                                                
                                                        }
                                                }
                                                
                                        }
                                }}
                                

                        }
                        sum++;
                        G=G-100*(ima+1);
                        p[ima]=p[ima]-1;
                        
                        
                        
                        if(p[ima]==0){
                                G=G-c[ima];
                                ima--;
                                
                                
                        }
                        
                        
                                
                        


                }
                System.out.println(sum);
                

                
                
                
                


                
        }
        
        
        

		
	
}