#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, pair<T1, T2> P) {
  return s << '<' << P.first << ", " << P.second << '>';
}
template <class T>
ostream& operator<<(ostream& s, vector<T> P) {
  for (int i = 0; i < P.size(); ++i) {
    if (i > 0) {
      s << " ";
    }
    s << P[i];
  }
  return s;
}
template <class T>
ostream& operator<<(ostream& s, vector<vector<T> > P) {
  for (int i = 0; i < P.size(); ++i) {
    s << endl << P[i];
  }
  return s << endl;
}
long long GCD(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return GCD(b, a % b);
}
inline long long mod(long long a, long long m) { return (a % m + m) % m; }
long long modmul(long long a, long long b, long long m) {
  a = mod(a, m);
  b = mod(b, m);
  if (b == 0) return 0;
  long long res = modmul(mod(a + a, m), b >> 1, m);
  if (b & 1) res = mod(res + a, m);
  return res;
}
long long modinv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a = mod(a - modmul(t, b, m), m);
    swap(a, b);
    u = mod(u - modmul(t, v, m), m);
    swap(u, v);
  }
  return mod(u, m);
}
pair<long long, long long> ChineseRem(const vector<long long>& b,
                                      const vector<long long>& m,
                                      long long lcm) {
  long long r = 0, M = 1;
  for (int i = 0; i < (int)b.size(); ++i) {
    long long d = GCD(M, m[i]);
    if ((b[i] - r) % d != 0) return make_pair(0, -1);
    long long p = modinv(M / d, m[i] / d);
    long long tmp = modmul((b[i] - r) / d, p, m[i] / d);
    r += modmul(M, tmp, lcm);
    M *= m[i] / d;
  }
  return make_pair(mod(r, M), M);
}
bool solve(long long N, long long M, long long K, vector<long long> A) {
  long long lcm = 1;
  for (int i = 0; i < K; ++i) {
    long long g = GCD(lcm, A[i]);
    lcm *= (A[i] / g);
    if (lcm > N || lcm < 0) return false;
  }
  vector<long long> b(K), m(K);
  for (int i = 0; i < K; ++i) b[i] = mod(-i, A[i]), m[i] = A[i];
  pair<long long, long long> res = ChineseRem(b, m, lcm);
  if (res.second == -1) return false;
  if (res.first == 0) res.first = lcm;
  if (res.first > M - K + 1) return false;
  for (int k = 0; k < K; ++k) {
    if (GCD(lcm, res.first + k) != A[k]) return false;
  }
  return true;
}
int main() {
  long long N, M, K;
  cin >> N >> M >> K;
  vector<long long> A(K);
  for (int i = 0; i < K; ++i) cin >> A[i];
  if (solve(N, M, K, A))
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}
