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;
void fact(vector<pair<long long, long long>> &v, long long k) {
  long long i;
  for (i = 1; i * i <= k; i++) {
    if (k % i == 0) {
      v.push_back(make_pair(i, k / i));
      if (k / i != i) {
        v.push_back(make_pair(k / i, i));
      }
    }
  }
}
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  long long a[n], b[m];
  long long i, j, p;
  for (i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (i = 0; i < m; i++) {
    cin >> b[i];
  }
  vector<long long> on1, on2;
  for (i = 0; i < n; i++) {
    long long c = 0;
    while (a[i] == 1 && i < n) {
      i++;
      c++;
    }
    if (c) on1.push_back(c);
  }
  for (i = 0; i < m; i++) {
    long long c = 0;
    while (b[i] == 1 && i < m) {
      i++;
      c++;
    }
    if (c) on2.push_back(c);
  }
  long long sm = 0;
  vector<pair<long long, long long>> v;
  fact(v, k);
  long long sz = v.size();
  long long sz1, sz2;
  sz1 = on1.size();
  sz2 = on2.size();
  long long ans = 0;
  for (i = 0; i < sz; i++) {
    long long sm1 = 0;
    long long sm2 = 0;
    for (j = 0; j < sz1; j++) {
      if (on1[j] >= v[i].first) {
        sm1 = sm1 + (on1[j] - v[i].first + 1);
      }
    }
    for (j = 0; j < sz2; j++) {
      if (on2[j] >= v[i].second) {
        sm2 = sm2 + (on2[j] - v[i].second + 1);
      }
    }
    ans += sm1 * sm2;
  }
  cout << ans;
}
