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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

*******************************************************************************/

#include <iostream>
#include <vector>
#include <cmath>

using namespace std;


void swap(int * p,int* q)
{
    int temp=*p;
    *p=*q;
    *q=temp;
    return ;
}

void help(int* mini,int* ans,vector <int> &A,int n,vector <int> tot,int* summer,int* gg,vector <int> numpro)
{
    if(n==1)
    {
        *summer=0;
        *ans=0;
        //sol.push_back(A);
        int i=0;
        for(;i<A.size();i++)
        {
            (*summer) += tot[A[i]];
            (*ans) += numpro[A[i]];
            if(*summer>=*gg)
            {
                *summer -= tot[A[i]];
                (*ans) -= numpro[A[i]];
                
                int re = 1;
                while(re < numpro[A[i]]){
                    
                    *summer += (A[i]+1)*100;
                    
                    if(*summer>=*gg)
                    {
                        *ans  += re;
                        *mini = min(*ans,*mini);
                        break;
                    }                    
                    re++;
                }
                
                if( *summer < *gg ){
                    *ans  += re;
                    *mini = min(*ans,*mini);
                }
            }
        }
        
    }
    
    for (int i=0; i<n; i++)
    {
        //help(sol,A,n-1);
        help(mini,ans,A,n-1,tot,summer,gg,numpro);
 
        // if size is odd, swap first and last
        // element
        if (n%2==1)
            swap(&A[0], &A[n-1]);
 
        // If size is even, swap ith and last
        // element
        else
            swap(&A[i], &A[n-1]);
    }
}

int main()
{
    int D,G;
    vector <int> numpro;
    vector <int> score;
    cin >> D;
    cin>>G;
    for(int i=0;i<D;i++)
    {
        int temp;
        cin>>temp;
        numpro.push_back(temp);
        
        int temps;
        cin>>temps;
        score.push_back(temps);
    }
    
    vector <int> tot(D,0);
    for(int i=0;i<D;i++)
    {
        tot[i]=numpro[i]*(100*(i+1))+score[i];
    }
    vector <int> A(D);
    for(int i=0;i<D;i++)
    {
        A[i]=i;
    }
    int summ=0;
    int* summer=&summ;
    int* gg=&G;
    int answer=0;
    int* ans=&answer;
    int alex = 1000000;
    int* mini = &alex;
    
    help(mini,ans,A,A.size(),tot,summer,gg,numpro);
    
    cout<< *mini;
    
    return 0;
}