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;
bool prime[10000002];
bool isPrime(long long n) {
  if (n <= 1) return false;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
void sieve(vector<long long> &v) {
  memset(prime, true, 10000002);
  prime[0] = prime[1] = false;
  for (long long i = 2; i * i < 10000002; i++) {
    if (prime[i] == true) {
      for (long long j = i + i; j < 10000002; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 0; i < 10000002; i++) {
    if (prime[i] == true) {
      v.push_back(i);
    }
  }
}
void countDivisor(vector<long long> &v, long long n) {
  for (long long i = 1; i * i <= n; i++) {
    if (i * i == n) {
      v.push_back(i);
    } else if (n % i == 0) {
      v.push_back(i);
      v.push_back(n / i);
    }
  }
}
void primeFactorization(vector<long long> &v, long long n) {
  while (n % 2 == 0) {
    v.push_back(2);
    n /= 2;
  }
  for (long long i = 3; i * i <= n; i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) {
    v.push_back(n);
  }
}
bool isPalindrome(string s) {
  long long l = 0;
  long long h = s.size() - 1;
  while (l < h) {
    if (s[l++] != s[h--]) {
      return false;
    }
  }
  return true;
}
long long sumofDigit(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long fastPower(long long a, long long b) {
  if (b == 0)
    return 1;
  else
    return a * fastPower(a, b - 1);
}
long long fastpowerMOD(long long x, long long p, long long m) {
  if (p == 1) return x;
  if (p % 2 == 0)
    return fastpowerMOD((x % m) * (x % m), p / 2, m) % m;
  else
    return ((fastpowerMOD((x % m) * (x % m), p / 2, m) % m) * (x % m)) % m;
}
void solve() {
  long long a, b, c;
  cin >> a >> b >> c;
  for (long long i = 0; i <= 1000; i++) {
    long long x = c - b * i;
    if (x < 0) continue;
    if (x % a == 0) {
      cout << "YES" << endl;
      return;
    }
  }
  cout << "NO" << endl;
}
signed main() {
  long long t = 1;
  while (t--) {
    solve();
  }
}