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;
typedef long long ll;
int n,m,ans,u[20];
int c[20],p[20];
void dfs(int x) {
	if (x>n) {
		ll s=0; int a=0;
		for (int i=1;i<=n;i++)
			if (u[i]) s+=p[i]+c[i]*i*100,a+=c[i];
		for (int i=n;i>=1;i--)
			if (!u[i]&&s<m) {
				if (s+(c[i]-1)*i*100<=m) s+=(c[i]-1)*i*100,a+=c[i]-1;
				else {
					int tmp=(m-s)/(i*100);
					a+=tmp; s+=tmp*i*100;
				}
			}
		if (s>=m) ans=min(ans,a);
		return;
	}
	u[x]=1; dfs(x+1); u[x]=0; dfs(x+1);
}
int main() {
	scanf("%d %d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%d %d",&c[i],&p[i]);
	ans=1e9;
	dfs(1);
	printf("%d\n",ans);
	return 0;
}