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 <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <bitset>
using namespace std;

typedef pair<int, int> P;
#define rep(i, n) for (int i=0; i<(n); i++)
#define all(c) (c).begin(), (c).end()
#define uniq(c) c.erase(unique(all(c)), (c).end())
#define index(xs, x) (int)(lower_bound(all(xs), x) - xs.begin())
#define _1 first
#define _2 second
#define pb push_back
#define INF 1145141919
#define MOD 1000000007

int N, M, Q;
int A[100000];
bool used[100000];
int perm[100000], rev[100000];
int C[100001];

signed main() {
  ios::sync_with_stdio(false); cin.tie(0);
  cin >> N >> M >> Q;
  rep(i, Q) cin >> A[i], A[i]--;

  C[0] = N;
  int top = 0;
  for (int i=Q-1; i>=0; i--) {
    int a = A[i];
    if (used[a]) {
      if (C[rev[a]] > 0) C[rev[a]+1]++, C[rev[a]]--;
    }
    else {
      used[a] = true;
      perm[top] = a;
      rev[a] = top;
      C[top]--, C[top+1]++;
      top++;
    }
  }
  vector<int> result;
  rep(i, M) used[i] = false;
  rep(i, M) {
    if (C[i]) break;
    result.pb(perm[i]);
    used[perm[i]] = true;
  }
  rep(i, M) if (!used[i]) result.pb(i);
  //cout<<"result: [";for (int x:result)cout<<x<<",";cout<<"]\n";
  bool bad = false;
  rep(i, M) {
    if (result[i] != perm[i]) bad = true;

    if (C[i+1] > 0 && bad) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
  return 0;
}
