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>
#include <stdlib.h>
#define MAX 1e9
using namespace std;
int main(){
    int d,g;
    cin>>d>>g;
    int p[d],c[d],i;
    for(i=0;i<d;i++) cin>>p[i]>>c[i];
    
    int ans=MAX,j;
    for(j=0; j<(1<<d);j++){
        int s=0,num=0,r_max=-1;
        for(i=0;i<d;i++){
            if(j>>i&1){
                s+= 100*(i+1)*p[i]+c[i];
                num+=p[i];
            }
            else r_max=i;
        }
        int s1,need;
        if(s<g){
            s1=100*(r_max+1);
            need=(g-s+s1-1)/s1;
            if(need>=p[r_max]) continue;
            num+=need;
        }
        ans=min(ans,num);
    }
    cout<<ans<<endl;
}
