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.Text; 
using System.Collections.Generic;
using System.Linq; 
using E = System.Linq.Enumerable; 

class Program {
     static int[] Quest;
     static int G;
     static int D;
     static int ans;
     static bool ok;
     static void Main(string[] args) {
     //入力を受け取る
     var DG = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
     
     D = DG[0];
     G = DG[1];
     
     Quest = new int[D+1];
     var bonuses = new int[D+1];
     
     for(int i = 1;i <= D ;i++){
         var PC = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
         
         var P = PC[0];
         var C = PC[1];
         Quest[i]   = P;
         bonuses[i] = C;
     }
     
     ans = int.MaxValue;
     for(int i = 0;i < Math.Pow(2,D);i++){
         ok = true;
         var point = 0;
         var step  = 0;
         //フラグが立っていればステップとボーナスを加算
         for(int j=0;j < D;j++){
             if((i&(int)Math.Pow(2,j)) != 0){
                 point += 100*(j+1)*Quest[j+1]+bonuses[j+1];
                 step  += Quest[j+1];
             }
         }
      //   Console.WriteLine(step+" "+point);
         //フラグ情報とポイントを渡し、残りのステップを計算
         step += AddRestStep(i,point);
         
        //Console.WriteLine(step+" "+point);
    
         if(ok && ans > step){
             ans = step;
         }
         
     }
     
     
        Console.WriteLine(ans);
     }
     
     static int AddRestStep(int flag,int point){
         var step = 0;
         
         for(int j = D;G > point && j > 0;j--){
             //ボーナスフラグが立っていなければ
             if((flag&(int)Math.Pow(2,j-1)) == 0){
                var comp = point+100*j*(Quest[j]-1);
                if(comp < G){
                    point = comp;
                    step  += Quest[j]-1;
                }else{
                    step  +=　(G-point-100*j)/(100*j)+1;
                    point = G;
                    break;
                }         
             }
         }
         
         if(point >= G){
             return step;
         }else{
             ok = false;
             return step;
         }
         
     }
}