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.Globalization;
using System.Linq;
using System.Runtime.Serialization;

namespace ABC104C
{
    internal class Program
    {
        public static void Main(string[] args)
        {
            int[] dg = ConsoleInput.IntArray();
            int D = dg[0];
            int G = dg[1];
            int[][] pc = new int[D][];
            for (int i = 0; i < D; i++)
            {
                pc[i] = ConsoleInput.IntArray();
            }

            int ap = Convert.ToInt32(Math.Pow(2, D));
            int[] pat = new int[ap];
            int[] patc = new int[ap];
            for (int i = 0; i < ap; i++)
            {
                pat[i] = 0;
                for (int j = 0; j < D; j++)
                {
                    if ((1 & i >> j) == 0)
                    {
                        continue;
                    }

                    pat[i] += pc[j][0] * (j + 1) * 100 + pc[j][1];
                    patc[i] += pc[j][0];
                }
            }

            int min = patc.Max();
            for (int i = 0; i < ap; i++)
            {
                for (int j = D - 1; j >= 0; j--)
                {
                    if ((1 & i >> j) == 0)
                    {
                        int max = pc[j][0] * (j + 1) * 100 + pat[i];

                        if (G < pat[i])
                        {
                            min = Math.Min(patc[i], min);
                            break;
                        }

                        if (max < G)
                        {
                            break;
                        }

                        int value = patc[i] + (G - pat[i]) / (j + 1) / 100 +
                                    ((G - pat[i]) % ((j + 1) * 100) > 0 ? 1 : 0);
                        min = Math.Min(value, min);
                        break;
                    }
                }
            }

            Console.WriteLine(min);
        }
    }

    internal static class ConsoleInput
    {
        public static string String()
        {
            return Console.ReadLine();
        }

        public static int Int()
        {
            return int.Parse(Console.ReadLine());
        }

        public static int[] IntArray()
        {
            return Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
        }

        public static long Long()
        {
            return long.Parse(Console.ReadLine());
        }

        public static long[] LongArray()
        {
            return Console.ReadLine().Split(' ').Select(long.Parse).ToArray();
        }

        public static double[] DoubleArray()
        {
            return Console.ReadLine().Split(' ').Select(double.Parse).ToArray();
        }
    }
}