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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#include <math.h>
#include <iomanip>
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF=1001001001;
vector<pair<int64_t,int64_t>>prime_factorize(int64_t x){
  vector<pair<int64_t,int64_t>>p;
  for(int64_t i=2;i*i<=x;i++){
    int cnt=0;
    if(x%i==0){
    while(x%i==0){cnt++;x/=i;}
    p.push_back(make_pair(i,cnt));
    }
  }
  if(x!=1){p.push_back(make_pair(x,1));}
  return p; 
}
int D,G;
int p[15];int c[15];
int ans=999999;
void score(vector<int> &A){
  int sum=0,cnt=0;//cout<<D<<" "<<G<<endl;
  for(int i=0;i<D;i++){
    sum+=(i+1)*100*A[i]+A[i]/p[i]*c[i];
    cnt+=A[i];//cout<<A[i]<<endl;
  }
  if(sum>=G){chmin(ans,cnt);}
  return;
}

void dfs(vector<int> &A,int i){
  if(A.size()==D){
    score(A);
    return ;
  }
  for(int v=0;v<=p[i];v++){
    A.push_back(v);
    dfs(A,i+1);
    A.pop_back();
  }
}
int main() {
  cin>>D>>G;
  for(int i=0;i<D;i++){
    cin>>p[i]>>c[i];
  }
  vector<int>A;
  dfs(A,0);
  cout<<ans<<endl;
  return 0;
}