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

namespace ABC
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] s = Console.ReadLine().Split(' ');
            int D = int.Parse(s[0]);
            int G = int.Parse(s[1]);
            List<int[]> list = new List<int[]>();
            for(int i = 0; i < D; i++)
            {
                int[] xz = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
                list.Add(xz);
            }

            int min = int.MaxValue;
            int n = (int)Math.Pow(2, D);
            for (int i = 0; i < n; i++)
            {
                int k = i, counter = 0, countProblems = 0;
                int sum = 0;
                List<int> plist = new List<int>();
                while (counter < D)
                {
                    int[] v = list[counter];
                    if (k % 2 == 1)
                    {
                        sum += v[0] * (counter + 1) * 100 + v[1];
                        countProblems += v[0];
                    }
                    else
                    {
                        plist.AddRange(Enumerable.Repeat((counter + 1) * 100, v[0] - 1));
                    }
                    k /= 2;
                    counter++;

                }
                plist.Reverse();
                if (sum < G)
                {
                    foreach (int p in plist)
                    {
                        sum += p; countProblems++;
                        if (sum >= G) break;
                    }
                }
                if (sum >= G && countProblems < min) min = countProblems;
            }
            Console.WriteLine(min);

        }
    }
}