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 <bits/stdc++.h>
using namespace std;

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef pair<LL, LL> PLL;

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N, M; cin >> N >> M;
  int Q; cin >> Q;
  VI xs(Q);
  REP(i,Q){
	cin >> xs[i];
	--xs[i];
  }

  vector<bool> use(M);
  VI as;
  for(int i=Q-1;i>=0;--i)
	if(!use[xs[i]]){
	  use[xs[i]] = true;
	  as.PB(xs[i]);
	}
  REP(i,M)
	if(!use[i])
	  as.PB(i);
  //REP(i,M) cout<<as[i]<<" ";cout<<endl;
  VI len(M+1);
  len[0] = N;
  VI idx(M);
  REP(i,M) idx[as[i]] = i;
  for(int i=Q-1;i>=0;--i){
	int x = xs[i];
	if(len[idx[x]]){
	  len[idx[x]+1]++;
	  len[idx[x]]--;
	}
  }
  REP(i,M)
	len[idx[i]+1] += len[idx[i]];
  
  cout << (len[M]==N?"Yes":"No") << endl;
  
  return 0;
}
