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.Scanner;
 
public class Main {
	public static int pow(int a, int b){
		int x=1;
		while(b>0){
			if((b&1)==1) x=(x*a);
			a=(a*a);
			b>>=1;
		}
		return x;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int D=sc.nextInt();
		int G=sc.nextInt();
		
		int[] p=new int[D+1];
		int[] c=new int[D+1];
		int[] score=new int[D+1];
		
		for(int i=1;i<=D;i++){
			p[i]=sc.nextInt();
			c[i]=sc.nextInt();
			score[i]=100*i*p[i]+c[i];
		}
		
		int pattern=pow(2,D);
		int[] num=new int[pattern];
		int min= 10000;
		for(int i=0;i<pattern;i++){
			int g=G;
			int n=0;
			for(int j=1;j<=D;j++){
				if( ((i>>(j-1))&1)==1 ){
					num[i]+=p[j];
					g-=score[j];
				}
			}
			if(g>0){
				int take=D;
				while(((i>>(take-1))&1)==1) take--;
				if(g>100*take*(p[take]-1)){
					continue;
				}else{
					num[i]+= (g-1)/(100*take)+1;
				}
			}
			min=Math.min(min, num[i]);
		}
		System.out.println(min);
	}
}