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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;


public class MyProgram
{	
    static int min = 999999999;
    static int G = 0;
    static int D = 0;
    static int[] Q = new int[10];
    static int[] B = new int[10];
    
    public static void Main()
	{
	    int[] n = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
	    D = n[0];
	    G = n[1];
	    
	    for(int i = 0 ; i < D ; i++)
	    {
	        n = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
	        Q[i] = n[0];
	        B[i] = n[1];
	    }
	    string ss = "";
	    tansaku(ss);
	    Console.WriteLine(min);
	}
	
	public static void tansaku(string s)
	{
	    if(s.Length == D)
	    {
	        keisan(s);
	    }
	    else
	    {
	        tansaku(s+"1");
	        tansaku(s+"0");
	    }
	}
	
	public static void keisan(string s)
	{
	    char[] c = s.ToCharArray();
	    int cnt = 0;
	    int a = 0;
	    
	    for(int i = 0 ; i < D ; i++)
	    {
	        if(c[i] == '1')
	        {
	            a += (Q[i] * (i+1) * 100 ) + B[i];
	            cnt += Q[i];
	        }
	    }
	    
	    for(int i = D-1 ; i >= 0 ; i--)
	    {
	        if(c[i] == '0')
	        {
	            int aa = 0;
	            while(G > a && aa < Q[i])
	            {
	                a += ((i+1) * 100);
	                aa ++;
	                cnt++;
	            }
	            break;
	        }
	    }
	    if(min > cnt && G <= a)
	    {
	        min = cnt;
	    }
	}
}