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<cstring>

using namespace std;

int D;
unsigned long long G,v[1111],w[1111],sum,cnt,minn=1000000000000ull;
bool check[1111];

int main(){
	cin>>D>>G;
	for(int i=0;i<D;++i)cin>>v[i]>>w[i];
	for(int i=0;i<1<<D;++i){
		cnt=sum=0ll;
		memset(check,false,sizeof(check));
		for(int j=0;j<D;++j)if(i&(1<<j))sum+=v[j]*(j+1ull)*100ull+w[j],cnt+=v[j],check[j]=true;
		if(sum>=G)goto endd;
		for(int j=D-1;~j;--j){
			if(check[j])continue;
			for(int k=1;k<=v[j];++k){
				sum+=(j+1ull)*100ull;
				++cnt;
				if(sum>=G)goto endd;
			}
		}
		endd:
		minn=min(minn,cnt);
	}
	cout<<minn;
}