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<vector>
using namespace std;

int main(){
	int d, g, min = 0;
	cin >> d >> g;
	vector<int> p(d), c(d);
	for (int i=0; i < d; i++) {
		cin >> p[i] >> c[i];
		min += p[i];
	}
	for (int i=0; i<(1<<d) - 1; i++){
		int s = 0;
		int max0 = 0;
		int count = 0;
		for (int j=0; j<d; j++){
			if(i & 1<<j) {
				s += (j+1)* 100 * p[j] + c[j];
				count += p[j];
			} else {
				max0 = j;
			}
		}
		if(s < g && (p[max0] - 1) * (max0+1) * 100 + s >= g) {
			count += (g - s + (max0+1) * 100 - 1) / ((max0+1) * 100);
			s = g;
		}
		if(s >= g && count < min) min = count;
	}
	cout << min << endl;
}

