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 POW(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  if (b == 1) return a % mod;
  long long y = POW(a, b / 2, mod);
  if (b % 2 == 0)
    return (y * y) % mod;
  else
    return (((y * y) % mod) * a) % mod;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
const long long MOD = 1000000007;
const long long SIZE = 1000001;
const int INF = 0x3f3f3f3f;
const long long ll_INF = 0x3f3f3f3f3f3f3f3f;
const long long MAXN = numeric_limits<long long>::max();
const long long MAX = 90000000;
vector<long long> solve(vector<long long> v) {
  int n = v.size();
  vector<long long> x(n + 1);
  int i = 0;
  while (i < n) {
    if (v[i] == 0)
      i++;
    else {
      int j = i;
      while (j < n && v[j] == 1) j++;
      for (int k = 1; k <= j - i; k++) x[k] += j - i - k + 1;
      i = j;
    }
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k;
  cin >> n >> m >> k;
  vector<long long> a(n);
  vector<long long> b(m);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < m; i++) cin >> b[i];
  vector<long long> x = solve(a);
  vector<long long> y = solve(b);
  long long c = 0;
  for (int i = 1; i < x.size(); i++) {
    if (k % i == 0 && k / i <= m) c += x[i] * y[k / i];
  }
  cout << c;
}
