Problem: Dante is engaged in a fight with "The Savior". Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.

For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.

Input

The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.

Output

Print "Yes" (without quotes) if Dante can deal exactly c damage to the shield and "No" (without quotes) otherwise.

Examples

Input

4 6 15


Output

No


Input

3 2 7


Output

Yes


Input

6 11 6


Output

Yes

Note

In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const unsigned int M = 1000000007;
inline long long read() {
  char ch = getchar();
  long long nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  long long ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
long long gcd(long long a, long long b, long long& x, long long& y) {
  x = 1, y = 0;
  long long x1 = 0, y1 = 1, a1 = a, b1 = b;
  while (b1) {
    long long q = a1 / b1;
    tie(x, x1) = make_tuple(x1, x - q * x1);
    tie(y, y1) = make_tuple(y1, y - q * y1);
    tie(a1, b1) = make_tuple(b1, a1 - q * b1);
  }
  return a1;
}
vector<long long> findXandY(long long A, long long B, long long N) {
  long long x, y;
  long long k = gcd(A, B, x, y);
  if (N % k != 0) {
    return {-1, -1};
  }
  x = x * N / k;
  y = y * N / k;
  long long p = 0;
  long long x1 = x + (p * B / k);
  while (x1 < 0) {
    p++;
    x1 = x + (p * B / k);
  }
  while (x1 > 0) {
    p--;
    x1 = x + (p * B / k);
    if (x1 < 0) {
      p++;
      break;
    }
  }
  x = x + (p * B / k);
  long long y1 = y - (p * A / k);
  if (y1 >= 0) {
    return {x, y1};
  }
  return {-1, -1};
}
void solution() {
  long long a;
  cin >> a;
  long long b;
  cin >> b;
  long long c;
  cin >> c;
  vector<long long> ans = findXandY(a, b, c);
  if (ans[0] == -1) {
    cout << "No\n";
  } else {
    cout << "Yes\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solution();
  return 0;
}