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)

#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include<iomanip>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
using namespace std;
int d,g;
int res=1001;
int p[10];
int c[10];
void dfs(int depth,int num,int score,int flag)
{
  if(depth==d)
  {
    //cout<<"dfs"<<score<<" "<<num<<endl;
    int tnum=0,tscore=score;
    for(int i=d-1;i>=0;i--)
    {
      if(g<=tscore)break;
      if((flag>>i)%2)continue;
      if(g-tscore<=(p[i]-1)*(i+1)*100)
      {
        tnum+=(g-tscore-1)/((i+1)*100)+1;
        tscore+=((g-tscore-1)/((i+1)*100)+1)*(i+1)*100;
        break;
      }
      else
      {
        tnum+=p[i]-1;
        tscore+=(p[i]-1)*(i+1)*100;
      }

    }
    if(g>tscore)return;
    res=min(res,num+tnum);return;
  }
  dfs(depth+1,num+p[depth],score+p[depth]*(depth+1)*100+c[depth],flag+(1<<depth));
  dfs(depth+1,num,score,flag);
}
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin>>d>>g;
  rep(i,d)cin>>p[i]>>c[i];
  dfs(0,0,0,0);
  cout<<res;





  return 0;

}