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<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
#include<map>
#include<queue>
#include<deque>
#include<iomanip>
#include<tuple>
using namespace std;
typedef long long int LL;
typedef pair<int,int> P;
typedef pair<LL,int> LP;
const int INF=1<<30;
const LL MAX=1e9+7;

void array_show(int *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%d%c",array[i],(i!=array_n-1?middle:'\n'));
}
void array_show(LL *array,int array_n,char middle=' '){
	for(int i=0;i<array_n;i++)printf("%lld%c",array[i],(i!=array_n-1?middle:'\n'));
}

int t[100005];
map<int,int> m1;
vector<int> v1,v2;//after,count

int main(){
	int n,m,q;
	int i,j,k;
	int a,b,c;
	cin>>n>>m>>q;
	for(i=0;i<q;i++){
		cin>>t[i];
	}
	v2.push_back(n);
	for(i=q-1;i>=0;i--){
		if(m1.find(t[i])==m1.end()){
			m1[t[i]]=v2.size();
			v1.push_back(t[i]);
			v2.push_back(0);
		}
		b=m1[t[i]];
		if(v2[b-1]>0)v2[b]++,v2[b-1]--;
	}
	for(i=1;i<=m+1;i++){
		if(m1.find(i)!=m1.end())continue;
		m1[i]=v1.size();
		v1.push_back(i);
		break;
	}
	int s=v2.back();
	for(i=v1.size()-1;i>=0;i--){
		if(v1[i-1]>v1[i])break;
		s+=v2[i-1];
	}
	if(s==n)cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
}