There are N arrays. The length of each array is M and initially each array contains integers (1，2，...，M) in this order.

Mr. Takahashi has decided to perform Q operations on those N arrays. For the i-th (1≤i≤Q) time, he performs the following operation.

* Choose an arbitrary array from the N arrays and move the integer a_i (1≤a_i≤M) to the front of that array. For example, after performing the operation on a_i=2 and the array (5，4，3，2，1), this array becomes (2，5，4，3，1).



Mr. Takahashi wants to make N arrays exactly the same after performing the Q operations. Determine if it is possible or not.

Constraints

* 2≤N≤10^5
* 2≤M≤10^5
* 1≤Q≤10^5
* 1≤a_i≤M

Input

The input is given from Standard Input in the following format:


N M
Q
a_1 a_2 ... a_Q


Output

Print `Yes` if it is possible to make N arrays exactly the same after performing the Q operations. Otherwise, print `No`.

Examples

Input

2 2
3
2 1 2


Output

Yes


Input

3 2
3
2 1 2


Output

No


Input

2 3
3
3 2 1


Output

Yes


Input

3 3
6
1 2 2 3 3 3


Output

No
#include <cstdio>
#include <vector>
#include <algorithm>

bool solve(int n, int m, std::vector<int> a) {
  std::vector<int> order(m + 1, -1), seq;
  std::vector<int> len(n, 0), cnt(m + 1, 0);
  int idx = 0;
  for (auto &v: a) {
    if (order[v] == -1) order[v] = ++idx, seq.push_back(v);
    if (cnt[v] == n) continue;
    if (len[cnt[v]] == order[v] - 1) ++len[cnt[v]], ++cnt[v];
  }
  int mx = len.back();
  cnt.assign(m + 1, 0);
  for (int i = 0; i < mx; ++i) cnt[seq[i]]++;
  std::vector<int> u;
  for (int i = 1; i <= m; ++i) if (cnt[i] == 0) {
    u.push_back(i);
  }
  for (int i = mx; i < idx; ++i) {
    if (u[i - mx] != seq[i]) return false;
  }
  return true;
}

int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  std::vector<int> a(q);
  for (int i = 0; i < q; ++i) {
    scanf("%d", &a[i]);
  }
  std::reverse(a.begin(), a.end());
  if (solve(n, m, a)) puts("Yes");
  else puts("No");
  return 0;
}
