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 <math.h>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>
#include <queue>
#include <map>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <sstream>
#include <set>
using namespace std;

int main()
{
  int n, m;
  cin >> n >> m;
  int Q;
  cin >> Q;

  int len = 0;
  vector<int> order(m, -1);
  vector<int> natural(m);
  vector<int> freq(m + 1);
  freq[0] = n;

  vector<int> ac(Q);
  for (int iter=0; iter<Q; iter++) {
    cin >> ac[Q-1-iter];
    ac[Q-1-iter] --;
  }

  for (int iter=0; iter<Q; iter++) {
    int a = ac[iter];
//    printf("%d %d\n", a, order[a]);

    if (order[a] == -1) {
      order[a] = len;
      natural[len] = a;
      freq[len] --;
      len ++;
      freq[len] ++;
      continue;
    }
    if (freq[order[a]] == 0) continue;
    freq[order[a]] --;
    freq[order[a]+1] ++;
  }

  for (int i=0; i<m; i++)
    if (order[i] == -1) {
      natural[len++] = i;
    }
  int first = 0;
  for (int i=0; i<=m; i++)
    if (freq[i]) {
      first = i;
      break;
    }
  for (int i=first; i+1<m; i++)
    if (natural[i] > natural[i+1]) {
      cout << "No" << endl;
      return 0;
    }

  cout << "Yes" << endl;
  return 0;
}
