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


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        //var n = int.Parse(Console.ReadLine());
        int[] dg = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var d = dg[0];
        var g = dg[1];

        int[] p = new int[d];
        int[] c = new int[d];
        for (int i = 0; i < d; i++)
        {
            int[] pc = Console.ReadLine().Split().Select(int.Parse).ToArray();
            p[i] = pc[0];
            c[i] = pc[1];
        }

        //0(false) = 解かない
        //1(true)  = 完全に解く
        //bit探索(bit全探索)
        long ans = long.MaxValue;
        for (int x = 0; x < Math.Pow(2, d); x++)
        {
            var array = new bool[d];

            // 右からk桁目を取り出す
            for (int k = 0; k < d; k++)
            {
                var y = (x >> k) & 1;
                if (y == 1) array[k] = true;
            }

            long tmp = 0;
            long count1 = 0;
            long last = 0;
            for (int k = 0; k < d; k++)
            {
                if (array[k])
                {
                    tmp += (k + 1) * 100 * p[k] + c[k];
                    count1 += p[k];
                }
                else
                {
                    last = k;
                }
            }

            //g以下の場合は中途半端に解く
            long count2 = 0;
            while (tmp < g && count2 < p[last])
            {
                tmp += (last + 1) * 100;
                count2++;
            }

            if (tmp > d)
            {
                ans = Math.Min(ans, count1 + count2);
            }
        }

        // 出力
        Console.WriteLine(ans);
    }
}