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>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<climits>
#include<cstring>
using namespace std;
#define MAXN 100
int n;
int G,T[MAXN+5],c[MAXN+5];
int cnt,vis[MAXN+5],ans=1000000000;
void dfs(int x,int G,int step)
{
    for(int i=n;i>=1;i--)
    if(!vis[i])
    {
        int M=T[i]*100*i+c[i];
        if(G>M)vis[i]=1,dfs(x+1,G-M,step+T[i]),vis[i]=0;
        else
        {
            int p=(G+i*100-1)/(i*100);
            ans=min(ans,step+min(p,T[i]));
        }
    }
}
int main()
{
    scanf("%d%d",&n,&G);
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&T[i],&c[i]);
    }
    dfs(1,G,0);
    printf("%d",ans);
}
