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 <iostream>
using namespace std;
int main(void){
    // Your code here!
    int n,m;
    cin>>n>>m;
    int s[n],t[n];
    int ww=100000000;
    int num=0,pt=0,w,w2;
    for(int i=0;i<n;i++){
        cin>>s[i]>>t[i];
    }
    for(int i=0;i<(1<<n);i++){
        pt=0;
        num=0;
        w=0,w2=0;
        for(int j=0;j<n;j++){
            if(i & (1<<j)){
                pt+=100*(j+1)*(s[j])+t[j];
                num+=s[j];
            }
            else {w=s[j]-1;w2=j+1;}
        }
        if (m>pt+w*w2*100)continue;
        else if(m<=pt)ww=min(ww,num);
        else ww=min(ww,num+(m-pt-1)/(w2*100)+1);
    }
    cout<<ww<<endl;
}
