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>
using namespace std;
#define int long long
const int N=15;
int sum=1e9,n,m,p[N],c[N];
signed main()
{
	scanf("%d%d",&n,&m);
	for (int i=0;i<n;i++)scanf("%d%d",&p[i],&c[i]);
	for (int t=0;t<1<<n;t++)
	 {
	 	int num=0,ans=0;
	 	for (int i=0;i<n;i++)
	 	 if ((1<<i)&t)num+=c[i]+p[i]*100*(i+1),ans+=p[i];
	 	for (int i=n-1;i>=0;i--)
		 if (num<m&&!((1<<i)&t))
		  {
		  	int s=min(p[i]-1,(m-num-1)/(100*(i+1))+1);
		  	num+=s*100*(i+1);
		  	ans+=s;
		  } 
		if (m<=num)sum=min(sum,ans);  
	 }
	printf("%d\n",sum); 
	return 0; 
} 