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 <string>
#include <vector>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<functional>
#include<list>
#include<deque>
#include<bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<cstring>
#include<sstream>
#include<complex>
#include<iomanip>
#include<numeric>
#include<cassert>
#define X first
#define Y second
#define pb push_back
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define repe(X,Y) for ((X) = 0;(X) < (Y);++(X))
#define peat(X,Y) for (;(X) < (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define eb emplace_back
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<class T> using vv=vector<vector<T>>;
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"{"; rep(i,t.size()) {os<<t[i]<<",";} os<<"}"<<endl; return os;}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}
const ll MOD=1e9+7;

int main(){
  ios_base::sync_with_stdio(false);
  cout<<fixed<<setprecision(0);
  int n,m,q;
  cin>>n>>m>>q;
  vector<int> a(q);
  rep(i,q){
    cin>>a[i]; --a[i];
  }
  reverse(all(a));
  vector<int> fin; fin.reserve(m);
  vector<int> inv(m,-1),cnt(m);
  for(int x:a){
    if(inv[x]<0){
      inv[x]=fin.size();
      fin.pb(x);
    }
    ++cnt[inv[x]];
    if(inv[x] && cnt[inv[x]-1]<cnt[inv[x]]){
      --cnt[inv[x]];
    }
  }
  rep(i,m) if(inv[i]<0){
    inv[i]=fin.size();
    fin.pb(i);
  }
  //cout<<cnt<<fin;
  rep(i,m-1) if(cnt[i]<n && fin[i]>fin[i+1]){
    cout<<"No"<<endl;
    return 0;
  }
  cout<<"Yes"<<endl;
  return 0;
}
