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-mcs 6.8.0.105) (judge verdict: Accepted)

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

/*
計算量について
n = 20くらいまでならbit全探索できるよ

なお、j番目のマスク（2^j）が欲しいだけなら、
「1 << j」として、
1をj回左シフトするだけで一瞬で作れるので、
わざわざリストを作る必要はない。
*/

/*
cf. Bit探索メモ
1. 2^N通りの全てを探索する場合
	0 <= i < 2^Nまでの全ての整数をチェックする必要。
	（なお、2^N = (1 << N)）

2. j番目のビットが立っているか確認する場合。
	対象の数値をiとしたとき、
	「(i & (1 << j)) == 0」ならば、ビットは立っていない。
	（AND演算子を使用。この時の「1 << j」（= 2^j）が、マスクになる。）

3. j番目のビットを立たせたい場合。
	対象の数値をiとしたとき、
	「k = i | (1 << j)」とすれば、
	kは、iのj番目のビットを立たせた数値となる。
	（OR演算子を使用。bit DPなどで役立つ。）
*/

class Program
{
	//Main
	static void Main(string[] args)
	{
		//読み込む
		string[] line = Console.ReadLine().Split(' ');
		int D = int.Parse(line[0]);
		int G = int.Parse(line[1]);
		int[] Pn = new int[D];
		int[] Cn = new int[D];
		for(int i=0; i<D; i++){
			string[] line1 = Console.ReadLine().Split(' ');
			Pn[i] = int.Parse(line1[0]);
			Cn[i] = int.Parse(line1[1]);
		}

		//最小問題数
		int ans = 10000000;

		//bit探索（0 <= 1 < 2^D）
		for(int i=0; i<(1<<D); i++){
			//得点合計
			int score = 0;
			//解いた問題数
			int cnt = 0;

			//各ビットをチェック
			/*
			ビットが立っている場合、コンプリートボーナスを得ている
			*/
			for(int j=0; j<D; j++){
				//ビットが立っていない場合、次へ
				if((i & (1 << j)) == 0){
					continue;
				}

				//更新
				score += (Pn[j]*(j+1)*100 + Cn[j]);
				cnt += Pn[j];
			}

			//得点が足りていない間、得点の高い問題から解き続ける
			for(int j=D-1; j>=0; j--){
				//得点が足りていたら、終わり
				if(score >= G){
					break;
				}

				//既に解いている場合（ビットが立っている）、スキップ
				if((i & (1 << j)) != 0){
					continue;
				}

				//解いていない場合、足りない分だけ解く
				int times = (G - score) / ((j+1)*100);
				times = Math.Min(times,Pn[j]);
				score += ((j+1)*100 * times);
				cnt += times;
			}

			//得点が足りていなかったら、次へ
			if(score < G){
				continue;
			}

			//最小回数を更新
			if(ans > cnt){
				ans = cnt;
			}
		}

		//出力
		Console.WriteLine(ans);
	}
}