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;

class Program
{
    static int[] p;
    static int[] c;
    static int G;
    static int D;

    static void Main(string[] args)
    {
        var dg = Console.ReadLine().Split(' ');
        D = int.Parse(dg[0]);
        G = int.Parse(dg[1]);
        p = new int[D];
        c = new int[D];
        for (int i = 0; i < D; i++)
        {
            var s = Console.ReadLine().Split(' ');
            p[i] = int.Parse(s[0]);
            c[i] = int.Parse(s[1]);
        }

        var ans = dfs(0, new bool[D]);
        Console.WriteLine(ans);
    }

    static int dfs(int n, bool[] v)
    {
        if (v.Length <= n)
        {
            return m1(v);
        }

        var left = dfs(n + 1, v);
        v[n] = true;
        var right = dfs(n + 1, v);
        v[n] = false;
        return Math.Min(left, right);
    }

    static int m1(bool[] v)
    {
        var g = 0;
        var ans = 0;

        for (int i = 0; i < D; i++)
        {
            if (v[i])
            {
                ans += p[i];
                g += (i + 1) * 100 * p[i];
                g += c[i];
            }
        }

        if (G <= g)
        {
            return ans;
        }

        for (int i = D - 1; i >= 0; i--)
        {
            if (!v[i])
            {
                var n = Math.Min(
                    p[i],
                    ceil(G - g, (i + 1) * 100));
                ans += n;
                g += (i + 1) * 100 * n;
            }
        }

        if (G <= g)
        {
            return ans;
        }
        else
        {
            return int.MaxValue;
        }
    }

    static int ceil(int a, int b)
    {
        return a / b + (a % b > 0 ? 1 : 0);
    }
}