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 rep(i, n) for (int i = 0; i < (n); ++i)
int p[15];
int c[15];

int main() {
	
	
	int d,g;cin>>d>>g;
	
	rep(i,d) cin>>p[i]>>c[i];
	
	int ans = 123456789;
	
	rep(i,(1<<d))
	{
		int score = 0;
		int num = 0;
		rep(j,d)if(i&(1<<j))
		{
			score += (j+1)*100*p[j] + c[j];
			num += p[j];
		}
		
		int ma;
		for(int j=d-1;j>=0;j--)if(!(i&(1<<j))) {ma=j;break;}
		if(score + (p[ma]-1)*100*(ma+1) < g)continue;
		
		int yozyo = score>=g?0:(g-score + 100*(ma+1) - 1) / (100*(ma+1));
	
		ans = min(ans,num+yozyo);
	}
	
	cout << ans << endl;
	
	
	
}