#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
const int mod = 1073741824;
vector<int> g[N];
int dp[N][N];
int gcd(int a, int b) {
  if (b == 0) return a;
  if (dp[a][b] != -1) return dp[a][b];
  return dp[a][b] = gcd(b, a % b);
}
int mb[N];
int pri[N], tot = 0;
bool vis[N];
void init() {
  memset(vis, false, sizeof vis);
  mb[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      pri[tot++] = i;
      mb[i] = -1;
    }
    for (int j = 0; j < tot && i * pri[j] < N; j++) {
      vis[i * pri[j]] = true;
      if (i % pri[j])
        mb[i * pri[j]] = -mb[i];
      else {
        mb[i * pri[j]] = 0;
        break;
      }
    }
  }
  memset(dp, -1, sizeof dp);
  for (int i = 1; i < N; i++) {
    for (int j = 1; j < N; j++)
      if (gcd(i, j) == 1) g[i].push_back(j);
  }
}
int main() {
  int a, b, c;
  init();
  scanf("%d%d%d", &a, &b, &c);
  long long ans = 0;
  int up = min(b, c);
  for (int i = 1; i <= a; i++) {
    int n = g[i].size();
    long long sum = 0;
    for (int ii = 0; ii < n && g[i][ii] <= up; ii++) {
      int d = g[i][ii];
      if (mb[d] == 0) continue;
      long long t1 = 0;
      int up1 = b / d;
      long long tmp = mb[d];
      for (int jj = 0; jj < n && g[i][jj] <= up1; jj++)
        t1 = (t1 + up1 / g[i][jj]) % mod;
      tmp = tmp * t1 % mod;
      int up2 = c / d;
      long long t2 = 0;
      for (int kk = 0; kk < n && g[i][kk] <= up2; kk++)
        t2 = (t2 + up2 / g[i][kk]) % mod;
      tmp = tmp * t2 % mod;
      sum = (sum + tmp) % mod;
    }
    ans = (ans + a / i * sum) % mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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:

<image>

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
const int mod = 1073741824;
vector<int> g[N];
int dp[N][N];
int gcd(int a, int b) {
  if (b == 0) return a;
  if (dp[a][b] != -1) return dp[a][b];
  return dp[a][b] = gcd(b, a % b);
}
int mb[N];
int pri[N], tot = 0;
bool vis[N];
void init() {
  memset(vis, false, sizeof vis);
  mb[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      pri[tot++] = i;
      mb[i] = -1;
    }
    for (int j = 0; j < tot && i * pri[j] < N; j++) {
      vis[i * pri[j]] = true;
      if (i % pri[j])
        mb[i * pri[j]] = -mb[i];
      else {
        mb[i * pri[j]] = 0;
        break;
      }
    }
  }
  memset(dp, -1, sizeof dp);
  for (int i = 1; i < N; i++) {
    for (int j = 1; j < N; j++)
      if (gcd(i, j) == 1) g[i].push_back(j);
  }
}
int main() {
  int a, b, c;
  init();
  scanf("%d%d%d", &a, &b, &c);
  long long ans = 0;
  int up = min(b, c);
  for (int i = 1; i <= a; i++) {
    int n = g[i].size();
    long long sum = 0;
    for (int ii = 0; ii < n && g[i][ii] <= up; ii++) {
      int d = g[i][ii];
      if (mb[d] == 0) continue;
      long long t1 = 0;
      int up1 = b / d;
      long long tmp = mb[d];
      for (int jj = 0; jj < n && g[i][jj] <= up1; jj++)
        t1 = (t1 + up1 / g[i][jj]) % mod;
      tmp = tmp * t1 % mod;
      int up2 = c / d;
      long long t2 = 0;
      for (int kk = 0; kk < n && g[i][kk] <= up2; kk++)
        t2 = (t2 + up2 / g[i][kk]) % mod;
      tmp = tmp * t2 % mod;
      sum = (sum + tmp) % mod;
    }
    ans = (ans + a / i * sum) % mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

```