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 <vector>
#include <algorithm>
using namespace std;
int main(){
  int n, m;
  cin >> n >> m;
  int q;
  cin >> q;
  vector<int> a(q);
  for(int i = 0; i < q; i++){
    cin >> a[i];
    a[i]--;
  }

  vector<int> b(m, -1);
  vector<int> pos(m, -1);
  vector<int> freq(m + 1,0);
  freq[0] = n;
  freq[m] = 0;
  int known = 0;
  for(int it = q - 1; it >= 0; it--){
    int i = a[it];
    if(pos[i] == -1){
      b[known] = i;
      pos[i] = known;
      freq[known]--;
      known++;
      freq[known]++;
      continue;
    }
    if(freq[pos[i]] > 0){
      freq[pos[i]]--;
      freq[pos[i] + 1]++;
    }
  }
  
  int first = -1;
  for(int i = 0; i <= m; i++){
    if(freq[i]  > 0){
        first = i;
        break;
    }
  }
  
  vector<int> all;
  for(int i = first; i < known; i++){
     all.push_back(b[i]);
  }
  for(int i = 0; i < m; i++){
    if(pos[i] == -1){
      all.push_back(i);
    }
  }
  
 sort(all.begin(), all.end());
  for(int i = first; i < known; i++){
    if(all[i - first] != b[i]){
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
}