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.Diagnostics;
using System.Linq;
using System.Text;

namespace bla
{
    class Program
    {
        static void Main()
        {
            string[] s = Console.ReadLine().Split();
            int d = int.Parse(s[0]);
            int g = int.Parse(s[1]) / 100;

            List<int> p = new List<int>();
            List<int> c = new List<int>();
            for (int i = 0; i < d; i++)
            {
                s = Console.ReadLine().Split();
                p.Add(int.Parse(s[0]));
                c.Add(int.Parse(s[1]) / 100);
            }
            while(p.Count < 10)
            {
                p.Add(0);
                c.Add(0);
            }

            int best = int.MaxValue;
            for (int i = 0; i < (1 << 10); i++)
            {
                int sum = 0;
                int count = 0;
                for (int j = 0; j < 10; j++)
                {
                    if ((i & (1 << j)) != 0)
                    {
                        sum += p[j] * (j+1) + c[j];
                        count += p[j];
                    }
                }
                if (sum < g)
                    continue;
                best = Math.Min(best, count);
                for (int j = 0; j < 10; j++)
                {
                    if ((i & (1 << j)) == 0)
                        continue;
                    int nsum = sum;
                    int ncount = count;
                    nsum -= c[j];
                    int r = 0;
                    while(nsum >= g && r <= p[j])
                    {
                        nsum -= j+1;
                        ncount--;
                        r++;
                    }
                    nsum += j + 1;
                    ncount++;
                    if(nsum >= g)
                        best = Math.Min(best, ncount);
                }
            }
            Console.WriteLine(best);
        }
    }
}