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



class Program
{
    const int MOD = 1000000007;
    const int INF = 1 << 30;

    static void Main(string[] args)
    {
        Solve();
    }

    static long GCD(long x,long y)
    {
        if (y == 0) return x;
        return GCD(y, x % y);
    }

    static long LCM(long x,long y)
    {
        return x * y / GCD(x, y);
    }

    static int keta(int x)
    {
        string s = x.ToString();
        return s[s.Length - 1]-'0';
    }

    static void Solve()
    {
        Scan sc = new Scan();
        write wr = new write();


        long ans = INF;
        var x = sc.intarr;
        int D = x[0];
        int G = x[1];
        int[] p = new int[D];
        int[] c = new int[D];
        for(int i = 0; i < D; ++i)
        {
            var y = sc.intarr;
            p[i] = y[0];
            c[i] = y[1];
        }

        for (int i = 0; i < (1 << D); ++i)
        {
            long sum = 0;
            long num = 0;

            for (int j = 0; j < D; ++j)
            {
                if ((i & (1 << j)) == 1)
                {
                    sum += c[j] + p[j] * 100 * (j + 1);
                    num += p[j];
                }
                
            }
            if (sum >= G) ans = Math.Min(ans, num);
            else
            {
                for(int j = D - 1; j >= 0; --j)
                {
                    if ((i & (1 << j)) == 1) continue;
                    for(int k = 0; k < p[j]; ++k)
                    {
                        if (sum >= G) break;
                        sum += 100 * (j + 1);
                        ++num;
                    }
                }
                ans = Math.Min(ans, num);
            }
        }
        wr.wri(ans);
        
    }


}
    




class Scan
{
        public string str => Console.ReadLine();

        public string[] strarr => str.Split(' ');

        public long[] longarr => strarr.Select(long.Parse).ToArray();
        public int[] intarr => strarr.Select(int.Parse).ToArray();
    public char[] chararr => str.ToArray();


}



class write
{
    public void wri<Type>(Type x)
    {
        Console.WriteLine(x);
    }
}