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 <iostream>
#include <set>
#include <vector>

using namespace std;

typedef long long ll;

void failure () {
  cout << "No" << endl;
  exit(0);
}

const int MAX_N = 1e5 + 5;

int arr [MAX_N];
set<int, greater<int>> pos [MAX_N];

int main () {
  ios::sync_with_stdio(false);

  int n, m, qc;
  cin >> n >> m >> qc;

  for (int i = 0; i < qc; i++) {
    cin >> arr[i];
    pos[arr[i]].insert(i);
  }

  set<int> seen;
  vector<int> subseq;
  for (int i = qc - 1; i >= 0; i--) {
    if (!seen.count(arr[i])) {
      subseq.push_back(arr[i]);
      pos[arr[i]].erase(i);
      seen.insert(arr[i]);
    }
  }

  int mn = m + 1;
  for (int i = 1; i <= m; i++) {
    if (!seen.count(i)) {
      mn = min(mn, i);
    }
  }

  while (!subseq.empty() && subseq.back() < mn) {
    mn = subseq.back();
    subseq.pop_back();
  }

  int K = subseq.size();
  if ((ll) K * (ll) n > qc) failure(); // prevent TLE

  for (int i = 0; i < n - 1; i++) {
    int cur = qc + 1;
    for (int u : subseq) {
      if (pos[u].upper_bound(cur) == pos[u].end()) failure();
      cur = *pos[u].upper_bound(cur);
      pos[u].erase(cur);
    }
  }

  cout << "Yes" << endl;
}
