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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main(){
  int d,g,m;
  m=pow(10,5);
  cin >> d >> g;
  vector<pair<int,int>> v;
  int a,b;
  for(int i=0;i<d;i++){
      cin >> a >> b;
      pair<int,int> p(a,b);
      v.push_back(p);
  }
  for(int i=0; i<(1<<d); i++){
    bitset<10> bit(i);
    bitset<10> check(0);
    int ans,item;
    ans=0;
    item=0;
    for(int j=0; j<d; j++){
      if(bit[j]==1){
        ans+=100*(j+1)*v.at(j).first+v.at(j).second;
        check.set(j,1);
        item+=v.at(j).first;
      }
    }
    int stop = d-1;
    while(true){
      if(ans>=g || stop<0){
        break;
      }
      if(check[stop]==0){
        check.set(stop,1);
        if(ans+100*(stop+1)*(v.at(stop).first-1)>g){
          item+=ceil((g-ans)/(100*(stop+1)));
          ans = g;
          break;
        }
        else{
          ans += 100*(stop+1)*(v.at(stop).first-1);
          item+=v.at(stop).first-1;
        }
      }
      stop--;
    }
    if(m>item && ans >= g){
      m=item;
    }
  }
  cout << m << endl;  
}