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 <stdio.h>
#include <cstdlib>
#include <cassert>
#include <iostream>
#include <set>
#include <vector>
#include <cstring>
#include <string>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <complex>
#include <map>
#include <queue>
#include <array>
#include <bitset>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<pii> vii;
typedef vector<string> vs;

void out(vi v) {
  for (auto x : v) cerr << x << ' ';
  cerr << endl;
}

int main() {
  int n, m, q;
  cin >> n >> m >> q;
  vi a(q);
  for (int i = 0; i < q; ++i) {
    cin >> a[i];
    --a[i];
  }
  vi was(m), ord;
  for (int i = a.size() - 1; i >= 0; --i) {
    if (!was[a[i]]) ord.push_back(a[i]);
    ++was[a[i]];
  }
  for (int i = 0; i < m; ++i) {
    if (!was[i]) ord.push_back(i);
    ++was[i];
  }
//  out(ord);
  bool ok = ord.size() == m;
  for (int i = 0; i < ord.size(); ++i) if (ord[i] != i) ok = 0;
  if (ok) {
    printf("Yes\n");
    return 0;
  }
  vi c(ord.size()), pos(m);
  for (int i = 0; i < ord.size(); ++i) pos[ord[i]] = i;
  for (int i = a.size() - 1; i >= 0; --i) {
    int x = pos[a[i]];
    if (x == 0) {
      ++c[0];
    } else {
      c[x] = min(c[x] + 1, c[x-1]);
    }
  }
//  out(c);
  for (int i = 0; i < m; ++i) {
    int x = pos[i];
    if (x == 0) {
      c[0] = n;
    } else {
      c[x] = max(c[x], c[x-1]);
    }
  }
//  out(c);
  for (int i = 0; i < c.size(); ++i) if (c[i] < n) {
    printf("No\n");
    return 0;
  }
  printf("Yes\n");
  return 0;
}