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;

const int maxn = 20, maxans = 1e3;
int d, g, p[maxn], c[maxn], ans;

int main(){
	cin >> d >> g;
	for (int i = 0; i < d; ++i)
		cin >> p[i] >> c[i];
	int mask = 1 << d;
	ans = maxans;
	for (int i = 0; i < mask; ++i) {
		int k = 0, sum = 0, ind = 0;
		for (int j = 0; j < d; ++j)
			if (i >> j & 1) {
				k += c[j] + p[j] * 100 * (j + 1);
				sum += p[j];
			}
			else 
				ind = j;
		if (k < g) {
			if (k + (p[ind] - 1) * 100 * (ind + 1) < g)
				continue;
			sum += ceil(double(g - k) / (100 * (ind + 1)));
		}
		ans = min(ans, sum);
	}
	cout << ans << '\n';
	return 0;
}
