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 <cstring>
#include <cassert>
#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

#define TRACE(x) cerr << #x << " = " << x << endl

typedef long long ll;
typedef pair<int, int> P;
#define X first
#define Y second

const int MAX = 1<<17;

int bio[MAX];
vector <int> Poz[MAX];
int p[MAX];
int nebitan[MAX];

void Nope()
{
  printf("No\n");
  exit(0);
}

vector <int> Kad(vector <int> V, int br)
{
  int ind = (int) V.size() - 1;
  vector <int> R;

  for (; ind >= 0 && Poz[br].size(); ) {
    if (Poz[br].back() < V[ind]) {
      R.push_back(Poz[br].back());
      ind--;
    }
    Poz[br].pop_back();
  }

  if (ind >= 0)
    Nope();

  reverse(R.begin(), R.end());

  return R;
}

int n, m, q;

int main()
{
  scanf("%d%d%d", &n, &m, &q);

  for (int i=0; i<q; i++) {
    scanf("%d", &p[i]);
    Poz[p[i]].push_back(i);
  }

  vector <int> T;
  for (int i=0; i<n; i++)
    T.push_back(MAX);

  vector <int> Por;
  for (int i=q-1; i>=0; i--) {
    if (bio[p[i]]++)
      continue;

    Por.push_back(p[i]);
  }

  for (int i=(int) Por.size()-1; i>=0; i--) {
    if (i != (int) Por.size() - 1 && Por[i] > Por[i+1])
      break;

    int da = 1;
    int poc = (i == (int) Por.size() - 1) ? 1 : Por[i+1];
    for (int j=poc; j<Por[i]; j++)
      if (!bio[j])
        da = 0;

    if (!da)
      break;

    nebitan[Por[i]] = 1;
  }

  memset(bio, 0, sizeof bio);

  for (int i=q-1; i>=0; i--) {
    if (bio[p[i]]++)
      continue;

    if (nebitan[p[i]])
      continue;

    T = Kad(T, p[i]);
  }

  printf("Yes\n");

  return 0;
}
