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.Linq;

namespace abc104_c
{
	class Program
	{
		static void Main(string[] args)
		{
			int[] x = Console.ReadLine().Split().Select(int.Parse).ToArray();
			Tuple<int, int>[] arr = new Tuple<int, int>[x[0]];
			for (int i = 0; i < x[0]; i++)
			{
				int[] a = Console.ReadLine().Split().Select(int.Parse).ToArray();
				arr[i] = new Tuple<int, int>(a[0], a[1]);
			}
			int ret = int.MaxValue;
			for (int i = 0; i < 2 << x[0]; i++)
			{
				int now = 0; int sum = 0;
				for (int j = 0; j < x[0]; j++)
				{
					if ((i & (1 << j)) > 0)
					{
						now += arr[j].Item1;
						sum += 100 * (j + 1) * arr[j].Item1 + arr[j].Item2;
					}
				}
				if (sum < x[1])
				{
					int j = x[0] - 1;
					while ((i & (1 << j)) > 0) j--;
					int point = 100 * (j + 1);
					int should = (x[1] - sum + (point - 1)) / point;
					if (should > arr[j].Item1) now += 999999;
					else now += (x[1] - sum + (point - 1)) / point;
				}
				ret = Math.Min(ret, now);
			}
			Console.WriteLine(ret);
		}
	}
}