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;

int main(){
	int d, g;
	cin >> d >> g;
	vector<int> p(d), c(d);
	for(int i=0; i < d; i++) {
		cin >> p.at(i) >> c.at(i);
	}

	int ans = 1 << 29;
	for(int i=0; i < (1 << d); i++) {
		int sum = 0, count = 0, last = -1;
		for(int j=0; j < d; j++) {
			if(i & (1 << j)) {
				sum += 100 * (j+1) * p.at(j) + c.at(j);
				count += p.at(j);
			}else{
				last = j;
			}
		}
		if(sum < g) {
			int point = 100 * (last + 1);
			int need =  (g - sum + (point-1)) / point;
			if(need >= p[last]) continue;
			count += need;
		}
		ans = min(ans, count);
	}
	cout << ans << endl;
}
