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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<55;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

int n,m,q;
vi a;

int main(){
	cin>>m>>n>>q;
	a=vi(q);
	for(int i=0;i<q;i++){
		cin>>a[i];
		a[i]--;
	}
	reverse(a.begin(),a.end());
	vi b(n,-1),c(n+1),b_(n);
	int id=0,M=0,B=1;
	c[0]=m;
	for(int i=0;i<q;i++) if(b[a[i]]==-1){
		b[a[i]]=id;
		b_[id++]=a[i];
	}
	for(int i=0;i<n;i++) if(b[i]==-1){
		b[i]=id;
		b_[id++]=i;
	}
	for(int i=0;i<q;i++){
		int I=b[a[i]];
		if(M<I) B=0;
		if(c[I]){
			c[I]--;
			c[I+1]++;
			M=max(M,I+1);
		}
	}
	int s=c[n]+c[n-1];
	for(int i=n-2;i>=0&&b_[i]<b_[i+1];i--) s+=c[i];
	cout<<(s==m&&B?"Yes":"No")<<endl;
}