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;

namespace ABC104C
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] bufs = Console.ReadLine().Split(' ');
            int D = Int32.Parse(bufs[0]);
            int G = Int32.Parse(bufs[1]);
            int[] p = new int[D];
            int[] c = new int[D];

            // int[] ave = new int[D];

            for (int i = 0; i < D; i++)
            {
                bufs = Console.ReadLine().Split(' ');
                p[i] = Int32.Parse(bufs[0]);
                c[i] = Int32.Parse(bufs[1]);
            }

            int res = 0;

            while (G > 0)
            {
                int max = 100;
                if(p[D - 1] * D * 100 >= G)
                {
                    max = G / (D * 100);
                    c[D - 1] = 0;
                    if (G % (D * 100) != 0) { max++; }
                }

                if (max < 100)
                {
                    for (int i = 0; i < D - 1; i++)
                    {
                        if (p[i] > max)
                        {
                            c[i] = 0;
                        }
                    }
                }

                int idx = 0;
                int ave = 0;
                for (int j = 0; j < D; j++)
                {
                    if (p[j] == 0) continue;
                    int tmp = (j + 1) * 100 + c[j] / p[j];
                    if (ave < tmp)
                    {
                        idx = j;
                        ave = tmp;
                    }
                }

                int unt = (idx + 1) * 100;
                int chk = G / unt;
                if (chk > p[idx])
                {
                    G -= p[idx] * unt + c[idx];
                    res += p[idx];
                    p[idx] = 0;
                    c[idx] = 0;
                }
                else
                {
                    res += chk;
                    if(G % unt != 0) { res++; }
                    G = 0;
                }

            }

            Console.WriteLine(res);
        }
    }
}