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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
#include <bits/stdc++.h>
typedef long long int ll;
using namespace std;
int main(){
ll d,g;
cin >>d>>g;
ll p[d],c[d],u[d];
for(ll i=0;i<d;i++) cin >> p[i]>>c[i];
ll ans=100000,sum,k,a;
for(ll i=0;i<1<<d;i++){
  priority_queue<ll> q;
  sum=0,k=0;
  for(ll j=0;j<d;j++) u[j]=0;
  for(ll j=0;j<d;j++){
    if(i&(1<<j)){
      sum+=((p[j]*100*(j+1))+c[j]);
      k+=p[j];u[j]=1;
    }
    else{
      for(ll k=0;k<p[j];k++){
        q.push((j+1)*100);
      }
    }
  }
  while(sum<g){
    if(q.empty()) break;
    sum+=q.top();
    k++;
    q.pop();
  }
  if(sum>=g) ans=min(ans,k);
}
cout <<ans<<endl;
}
