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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
#define ALL(a)  (a).begin(),(a).end()
const ll INF = 1e18+18;
const ll MAX = 200000;
const ll MOD = 1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;



int main(){
  ll D, G;
  cin>>D>>G;
  vector<ll> P(D), C(D);
  rep(i, D){
    cin>>P[i]>>C[i];
  }
  ll Ans=INF;
  for(ll i=0; i<(1<<D); i++){
    ll Point=0;
    ll Cou=0;
    rep(j, D){
      if(i&(1<<j)){
        Point+=C[j];
        Point+=(100*(j+1))*P[j];
        Cou+=P[j];
      }
    }
    //cout<<Point;
    if(Point>=G){
      Ans=min(Ans,Cou);
      continue;
    }
    //cout<<'&';
    for(ll j=D-1; j>=0; j--){
      //cout<<Point<<" "<<Cou<<endl;
      if(i&(1<<j))continue;
      ll K=(G-Point)/((j+1)*100);
      if((G-Point)%((j+1)*100)>0)K++;
      //cout<<K<<" "<<endl;
      if(K<=P[j]-1){
        Cou+=K;
        Point=INF;
        break;
      }else{
        Cou+=P[j]-1;
        Point+=((j+1)*100)*(P[j]-1);
      }
    }
    if(Point<G)Cou=INF;
	Ans=min(Ans, Cou);

  }
  cout<<Ans<<endl;
}