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: Wrong Answer)

using System;
using System.IO;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;
using System.Numerics;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Math;
using Debug = System.Diagnostics.Debug;
using LayoutKind = System.Runtime.InteropServices.LayoutKind;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;


static class P
{
    static void Main()
    {
        var dg = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var d = dg[0];
        var g = dg[1];
        var pc = Enumerable.Repeat(0, d).Select(_ => Console.ReadLine().Split().Select(int.Parse).ToArray()).ToArray();
        int res = int.MaxValue;
        for (int i = 0; i < (1 << d); i++)
        {
            int remain = g;
            int count = 0;
            int max = -1;
            for (int j = 0; j < d; j++)
            {
                if (((i >> j) & 1) == 1)
                {
                    remain -= (j + 1) * pc[j][0] * 100 + pc[j][1];
                    count += pc[j][0];
                }
                else max = j;
            }
            if (max < 0 || remain > ((max + 1) * 100) * pc[max][0]) continue;
            count += (int)Ceiling((double)Max(0, remain) / ((max + 1) * 100));
            res = Min(res, count);
        }
        Console.WriteLine(res);
    }
}