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;
 
int N,M,Q;
 
int b[100005],c[100005],d[100005],e[100005],f[100005]; bool g[100005],h[100005];

vector<int> a,v,x;
 
const int INF=1000000009;
 
int main(){
 
	cin>>N>>M>>Q;
 
	for(int i=0; i<Q; i++){
		int x;
		scanf("%d",&x);
		a.push_back(x);
		b[i]=x;
		c[x]++;
	}

	sort(b,b+Q); M=b[Q-1];

	int j=1;
	
	for(int i=Q-1; i>=0; i--){
		if(!g[a[i]]){
			d[j]=a[i]; g[a[i]]=1; j++;
		}
	}

	for(int i=1; i<=M; i++){
		if(g[i]==0){
			v.push_back(i);
		}
	}

	for(int i=j; i<=M; i++){
		d[i]=v[0]; v.erase(v.begin());
	}

	int esize=M; bool no=1;

	for(int i=M-1; i>=1; i--){
		if(d[i]>d[i+1]){
			esize=i; no=0;
			break;
		}
	}

	if(no){
		cout<<"Yes"<<endl; return 0;
	}

	int k=1;

	for(int i=esize; i>=1; i--){
		e[k]=d[i]; k++;
	}

	int count=0,time=0; bool find=0; 

 while(1){

	k=1; find=0; 

	for(int i=0; i<a.size(); i++){

	 if(a[i]==e[k]&&h[i]==0){
	 	h[i]=1;
	 	k++;
	 }
	 
	 if(k==esize+1){
		count++; k=1; find=1;
	 }

	}

	if(!find) break;
	if(count>=N) break;
		
 }

 //cout<<z<<endl;
 //cout<<count<<endl;

	if(count>=N) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;

	return 0;
 
}