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 N = 2010;
int pri[N], pri_cnt, mu[N];
bool mark[N];
void sieve() {
  int i, j;
  mu[1] = 1;
  for (i = 2; i < N; i++) {
    if (!mark[i]) {
      pri[++pri_cnt] = i;
      mu[i] = -1;
    }
    for (j = 1; j <= pri_cnt; j++) {
      int t = i * pri[j];
      if (t >= N) {
        break;
      }
      mark[t] = 1;
      if (i % pri[j] == 0) {
        mu[t] = 0;
        break;
      }
      mu[t] = -mu[i];
    }
  }
}
bool hav[N][N];
int G[N][N];
int gcd(int x, int y) {
  if (y == 0) {
    return x;
  }
  if (hav[x][y]) {
    return G[x][y];
  }
  G[x][y] = G[y][x] = gcd(y, x % y);
  hav[x][y] = hav[y][x] = 1;
  return G[x][y];
}
int main() {
  sieve();
  int a, b, c;
  scanf("%d%d%d", &a, &b, &c);
  if (b > c) {
    swap(b, c);
  }
  int x, y, z, t;
  unsigned int Ans = 0;
  for (x = 1; x <= a; x++) {
    unsigned int sum = 0;
    for (t = 1; t <= b; t++) {
      unsigned int s1 = 0, s2 = 0;
      for (y = t; y <= b; y += t) {
        if (gcd(x, y) == 1) {
          s1 += (b / y);
        }
      }
      for (z = t; z <= c; z += t) {
        if (gcd(x, z) == 1) {
          s2 += (c / z);
        }
      }
      sum += mu[t] * s1 * s2;
    }
    Ans += sum * (a / x);
  }
  printf("%u\n", Ans & 1073741823);
  return 0;
}