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.Linq;
using System.Text;

namespace AtCoder
{
	public class Program
	{
		// ABC104-C ng
		static void Main(string[] args)
		{
			string[] vs = Console.ReadLine().Split();
			int D = int.Parse(vs[0]);   // 種類		1～10
			int G = int.Parse(vs[1]);   // 目標点
			int[] count = new int[D];   // 問題数	1～100
			int[] basic = new int[D];   // 基本点
			int[] bonus = new int[D];   // 追加点	100～1,000,000
			for (int i = 0; i < D; i++) {
				vs = Console.ReadLine().Split();
				count[i] = int.Parse(vs[0]);
				basic[i] = (i + 1) * 100;
				bonus[i] = int.Parse(vs[1]);
			}

			int answer = int.MaxValue;
			Queue<Item> items = new Queue<Item>();
			HashSet<string> hashSet = new HashSet<string>();
			items.Enqueue(new Item(D));
			while (items.Count > 0) {
				Item item = items.Dequeue();
				if (item.Point >= G) {
					answer = Math.Min(answer, item.Answer);
				} else {
					for (int i = D - 1; i >= 0; i--) {
						if (item.Count[i] == 0) {
							int k = (G - item.Point) / basic[i] + ((G - item.Point) % basic[i] > 0 ? 1 : 0);
							if (k < count[i]) {
								answer = Math.Min(answer, item.Answer + k);
							}
							break;
						}
					}
				}

				for (int i = 0; i < D; i++) {
					if (item.Count[i] == 0) {
						Item tmp = new Item(item);
						tmp.Answer += count[i];
						tmp.Count[i] = count[i];
						tmp.Point += basic[i] * count[i] + bonus[i];
						if (!hashSet.Contains(tmp.ToString())) {
							items.Enqueue(tmp);
							hashSet.Add(tmp.ToString());
						}
					}
				}
			}
			Console.WriteLine(answer);
		}
	}

	public class Item
	{
		public int Answer = 0;
		public int[] Count;
		public int Point = 0;

		public Item(int c)
		{
			Count = new int[c];
		}

		public Item(Item item)
		{
			Answer = item.Answer;
			Count = (int[])item.Count.Clone();
			Point = item.Point;
		}

		public override string ToString()
		{
			return string.Join(",", Count.Select(p => p.ToString("D3")));
		}
	}

}