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;
using ll = long long int;
vector<ll> fun(vector<ll> v) {
  int n = v.size();
  vector<ll> ans(n + 1, 0);
  int i = 0;
  int len = 0;
  while (i < n) {
    if (v[i] == 0) {
      len = 0;
      i++;
      continue;
    }
    while (i < n && v[i] == 1) {
      len++;
      i++;
    }
    for (int j = 1; j <= len; j++) {
      ans[j] += len + 1 - j;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll n, m, k;
  cin >> n >> m >> k;
  vector<ll> a(n);
  vector<ll> b(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int j = 0; j < m; j++) cin >> b[j];
  auto a1 = fun(a);
  auto b1 = fun(b);
  ll x = sqrt(k);
  ll ans = 0;
  for (ll y = 1; y <= x; y++) {
    if (k % y == 0) {
      if (y <= n && k / y <= m) ans += a1[y] * b1[k / y];
      if (y <= m && k / y <= n && y != k / y) ans += a1[k / y] * b1[y];
    }
  }
  cout << ans;
  return 0;
}
