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<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
int ans[200000];
int now[200000];
int aaa[200000];
int main()
{
	int num,len,query;
	scanf("%d%d%d",&num,&len,&query);
	vector<int>v;
	for(int i=0;i<query;i++)
	{
		int z;
		scanf("%d",&z);
		z--;
		v.push_back(z);
	}
	reverse(v.begin(),v.end());
	set<int>se;
	vector<int>dd;
	for(int i=0;i<v.size();i++)
	{
		if(se.find(v[i])==se.end())
		{
			ans[v[i]]=se.size()+1;
			se.insert(v[i]);
			dd.push_back(v[i]);
		}
		int x=ans[v[i]];
		int beg=0,end=num;
		for(;;)
		{
			if(beg==end)break;
			int med=(beg+end)/2;
			if(now[med]<x)end=med;
			else beg=med+1;
		}
		//printf("  %d %d  %d\n",beg,v[i]+1,x);
		if(beg!=num&&now[beg]==x-1)
		{
			now[beg]++;
		}
	}
	for(int i=0;i<len;i++)if(ans[i]==0)dd.push_back(i);
	vector<int>z;
	for(int i=0;i<now[num-1];i++)
	{
		z.push_back(dd[i]);
		aaa[dd[i]]=1;
	}
	for(int i=0;i<len;i++)if(aaa[i]==0)z.push_back(i);
	//for(int i=0;i<len;i++)printf("%d %d\n",dd[i]+1,z[i]+1);
	if(dd==z)printf("Yes\n");
	else printf("No\n");
}