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 d, best = INT_MAX;
long long g, c[15], p[15];
bool u[15];

void go(int a, long long b) {
	for (int i = 1; i <= d; i++)
		if (!u[i]) {
			if (b + i * 100 * p[i] + c[i] < g) {
				u[i] = 1;
				go(a + p[i], b + i * 100 * p[i] + c[i]);
				u[i] = 0;
			} else if (b + i * 100 * p[i] < g)
				best = min((long long)best, a + p[i]);
			else
				best = min((long long)best, a + (g - b + i * 100 - 1) / (i * 100));
		}
}

int main() {
	scanf("%d%lld", &d, &g);
	for (int i = 1; i <= d; i++)
		scanf("%lld%lld", p + i, c + i);
	go(0, 0);
	printf("%d\n", best);
	return 0;
}
