You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.

How many subrectangles of size (area) k consisting only of ones are there in c?

A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.

The size (area) of a subrectangle is the total number of cells in it.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.

The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.

Output

Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.

Examples

Input


3 3 2
1 0 1
1 1 1


Output


4


Input


3 5 4
1 1 1
1 1 1 1 1


Output


14

Note

In first example matrix c is:

<image>

There are 4 subrectangles of size 2 consisting of only ones in it:

<image>

In second example matrix c is:

<image>
#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, k;
  cin >> n >> m >> k;
  long long int x[n];
  long long int y[m];
  cin >> x[0];
  for (long long int i = 1; i < n; i++) {
    cin >> x[i];
  }
  cin >> y[0];
  for (long long int i = 1; i < m; i++) {
    cin >> y[i];
  }
  vector<long long int> divis;
  for (long long int i = 1; i * i <= k; i++) {
    if (k % i == 0) {
      divis.push_back(i);
      if (k / i != i) divis.push_back(k / i);
    }
  }
  sort(divis.begin(), divis.end());
  long long int ans = 0;
  for (long long int i = 0; i < divis.size(); i++) {
    long long int c = divis[i];
    long long int d = k / c;
    long long int a = 0, b = 0;
    long long int cnt = 0;
    for (long long int j = 0; j < n; j++) {
      if (x[j] == 1)
        cnt++;
      else
        cnt = 0;
      if (cnt == c) {
        cnt--;
        a++;
      }
    }
    cnt = 0;
    for (long long int j = 0; j < m; j++) {
      if (y[j] == 1)
        cnt++;
      else
        cnt = 0;
      if (cnt == d) {
        cnt--;
        b++;
      }
    }
    ans += a * b;
  }
  cout << ans << endl;
}
