Problem: Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

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Find the sum modulo 1073741824 (230).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 2000).

Output

Print a single integer — the required sum modulo 1073741824 (230).

Examples

Input

2 2 2


Output

20


Input

4 4 4


Output

328


Input

10 10 10


Output

11536

Note

For the first example.

  * d(1·1·1) = d(1) = 1; 
  * d(1·1·2) = d(2) = 2; 
  * d(1·2·1) = d(2) = 2; 
  * d(1·2·2) = d(4) = 3; 
  * d(2·1·1) = d(2) = 2; 
  * d(2·1·2) = d(4) = 3; 
  * d(2·2·1) = d(4) = 3; 
  * d(2·2·2) = d(8) = 4. 



So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1 << 30;
vector<int> pr;
int pchk[2010];
void sieve() {
  for (int i = 2; i < 2010; i++) {
    if (!pchk[i]) {
      pr.push_back(i);
    }
    for (int j = 0; j < pr.size() && i * pr[j] < 2010; j++) {
      pchk[i * pr[j]] = 1;
      if (i % pr[j] == 0) break;
    }
  }
}
int a, b, c;
unordered_map<long long, int> mm[304];
int dp(int a, int b, int c, int p) {
  if (p == -1) return 1;
  long long mul = (2001LL * a + b) * 2001 + c;
  if (mm[p].find(mul) != mm[p].end()) return mm[p][mul];
  int &ret = mm[p][mul];
  ret = 0;
  int q = pr[p];
  for (int i = 1, x = 0; i <= a; i *= q, x++) {
    for (int j = 1, y = 0; j <= b; j *= q, y++) {
      for (int k = 1, z = 0; k <= c; k *= q, z++) {
        ret += 1LL * (x + y + z + 1) * dp(a / i, b / j, c / k, p - 1) % mod;
        ret %= mod;
      }
    }
  }
  return ret;
}
int main() {
  sieve();
  cin >> a >> b >> c;
  cout << dp(a, b, c, pr.size() - 1);
}