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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  bool a[n], b[m];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) cin >> b[i];
  vector<pair<long long, long long> > v;
  for (long long i = 1; i * i <= k; ++i) {
    if (k % i == 0) {
      v.push_back(make_pair(i, k / i));
    }
  }
  long long c = 0;
  vector<long long> c_a(n + 1, 0), c_b(m + 1, 0);
  for (long long i = 0; i < n; i++) {
    if (a[i]) {
      ++c;
      for (long long j = c; j > 0; --j) c_a[j]++;
    } else {
      c = 0;
    }
  }
  c = 0;
  for (long long i = 0; i < m; i++) {
    if (b[i]) {
      ++c;
      for (long long j = c; j > 0; --j) c_b[j]++;
    } else {
      c = 0;
    }
  }
  long long sm = 0;
  for (auto i : v) {
    if (i.first != i.second) {
      if (i.first <= n && i.second <= m) sm += c_a[i.first] * c_b[i.second];
      if (i.first <= m && i.second <= n) sm += c_a[i.second] * c_b[i.first];
    } else if (i.first <= n && i.second <= m)
      sm += c_a[i.first] * c_b[i.second];
  }
  cout << sm;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
