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 MAX = 1e10;
long long int mod = 998244353;
void seq(vector<long long int> &a, vector<long long int> &b) {
  long long int i = 0;
  long long int n = a.size();
  while (i < n) {
    if (a[i] != 1) {
      i++;
      continue;
    }
    long long int j = i;
    while (j < n && a[j] == 1) j++;
    long long int val = j - i;
    long long int x = 1;
    for (long long int k = val; k >= 1; k--) {
      b[k] += x;
      x++;
    }
    i = j;
  }
}
void divisor(vector<long long int> &v, long long int k) {
  long long int val = sqrt(k);
  for (long long int i = 1; i < val + 1; i++) {
    if (k % i == 0) {
      if (i * i == k)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(k / i);
      }
    }
  }
}
void solve() {
  long long int n, m, k;
  cin >> n >> m >> k;
  vector<long long int> a(n), b(m);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < m; i++) cin >> b[i];
  vector<long long int> cnta(n + 1), cntb(m + 1), div;
  seq(a, cnta);
  seq(b, cntb);
  divisor(div, k);
  long long int ans = 0;
  for (auto x : div) {
    if (x <= n && k / x <= m) ans += cnta[x] * cntb[k / x];
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  for (long long int i = 0; i < t; i++) {
    solve();
  }
}
