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;
int main() {
  int n, m, k;
  long long res = 0;
  cin >> n >> m >> k;
  map<int, long long> dp;
  bool arr[n + 1], s[m + 1];
  for (int i = 0; i < n; i++) cin >> arr[i];
  arr[n] = 0;
  for (int i = 0; i < m; i++) cin >> s[i];
  s[m] = 0;
  int ones = 0;
  for (int i = 0; i < n + 1; i++) {
    if (arr[i])
      ones++;
    else if (ones) {
      if (dp.find(ones) != dp.end()) {
        res += dp[ones];
        ones = 0;
        continue;
      }
      int tones = 0;
      long long dpres = 0;
      long long prev = 0;
      for (int j = 0; j < m + 1; j++) {
        if (s[j]) {
          tones++;
          dpres += prev;
          if (k % tones == 0 && tones * ones >= k) {
            dpres += ones - (k / tones) + 1;
            prev += ones - (k / tones) + 1;
          }
        } else {
          tones = 0;
          prev = 0;
        }
      }
      dp[ones] = dpres;
      res += dpres;
      ones = 0;
    }
  }
  cout << res;
  return 0;
}
