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: Accepted)

    //        VIVEK UPADHYAY
    //        MANIT, BHOPAL(M.P)


//                         JAB-TAK TODEGA NHI,
//                         TAB -TAK CHODEGA NHI :D
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<cstdlib>
#include<unistd.h>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<utility>
#include<cmath>
#include<set>
#include<unordered_map>
#include<cstdio>
#include<map>
#include<numeric>
using namespace std;
#define F first
#define S second
#define mk(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define LINE cout<<endl
#define ok cout<<"ok\n" ;
typedef long long ll;
const int MAXN = 1e5 + 10 ;
const int MAX = 1e7 ;
double pi = 3.1415926535897932384626433832795;
const ll inf=1e18 + 10 ;
ll mod = 1e9 + 7 ;
int main() {
    
    
    int D , G ;
    cin >> D>>G ;
    int a[D][2] ;
    for(int i=0;i<D;i++) cin>>a[i][0] >>a[i][1] ;
    
    int  mx = 10000 ;
    for(int mask=0;mask<(1<<D);mask++){
        int tot =0 ,cnt=0;
        for(int j=0;(1<<j)<=mask;j++)
            if(mask&(1<<j)){
                tot+=a[j][1] +(a[j][0])*100*(j+1);
                cnt+=a[j][0] ;
            }
        tot = G-tot ;
       // cout<<tot<<" "<<cnt<<endl;
        if(tot<=0){
            mx = min(mx , cnt) ;
            continue ;
        }
        
        for(int i=D-1;i>=0;i--){
            if(((1<<i)&mask)==0){
                
                if((a[i][0]-1)*(i+1)*100 >=tot){
                    
                    int v = ceil(tot*1.0/(100*(i+1))) ;
                    tot =0 ;
                    cnt+=v ;
                    
                }
                else{
                    tot-=(a[i][0]-1)*100*(i+1) ;
                    cnt+=a[i][0]-1 ;
                }
            }
            if(tot<=0) break ;
            
        }
        
        if(tot<=0)
            mx = min(cnt , mx) ;
          
    }
    cout<<mx<<endl;
    return 0 ;
}