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.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;

class Program
{
    static void Main()
    {
        int[] DG = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
        int D = DG[0];
        int G = DG[1];
        var PandC =
            Enumerable
            .Range(0, D)
            .Select(n => Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray())
            .Select(arr => new { p = arr[0], c = arr[1] })
            .ToArray();

        //手をつける難易度の順序のバリエーションは
        var patterns = MathUtil.GetPermutation(D - 1);

        int min = int.MaxValue;

        foreach(var pattern in patterns)
        {
            var cnt = 0;
            var point = 0;

            foreach(var x in pattern)
            {
                if (PandC[x].p * (x + 1) * 100 + point >= G)
                {
                    for (int i = 0; i < PandC[x].p; i++)
                    {
                        cnt++;
                        point += (x + 1) * 100;
                        if (i == PandC[x].p - 1) point += PandC[x].c;
                        if (point >= G) break;
                    }
                    if (point >= G) break;
                }
                else
                {
                    cnt += PandC[x].p;
                    point += (x + 1) * 100 * PandC[x].p + PandC[x].c;
                }
            }

            min = Math.Min(min, cnt);
        }

        Console.WriteLine(min);
    }
    
}



static class MathUtil
{
    public static IEnumerable<IEnumerable<int>> GetPermutation(int n)
    {
        if (n == 0)
        {
            yield return Single(0);
            yield break;
        }

        for (int i = 0; i <= n; i++)
        {
            foreach (var x in GetPermutation(n - 1))
            {
                yield return Single(i).Concat(x.Select(itm => itm < i ? itm : itm + 1));
            }
        }
    }

    private static IEnumerable<int> Single(int x)
    {
        yield return x;
    }

    public static IEnumerable<int> EnumeratePrimeNumbers()
    {
        List<int> primes = new List<int>() { 2 };
        yield return 2;

        for (int i = 3; i < int.MaxValue; i++)
        {
            bool isPrime = true;
            int Sqrt = (int)Math.Sqrt(i) + 1;

            foreach (var prime in primes)
            {
                if (prime > Sqrt)
                {
                    break;
                }
                if (i % prime == 0)
                {
                    isPrime = false;
                    break;
                }
            }

            if (isPrime)
            {
                primes.Add(i);
                yield return i;
            }
        }
    }
}