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<cmath>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;

int n,m;
int q;
int a[100011];
int b[100011];
int lenb;


bool has[100011];

int calc[100011];

int atob[100011];


int main()
{
	int i,j;
	
	scanf("%d %d",&n,&m);
	scanf("%d",&q);
	for(i=1;i<=q;i++)
	{
		scanf("%d",&a[i]);
	}
	memset(has,0,sizeof(has));
	memset(atob,0,sizeof(atob));
	lenb=0;
	for(i=q;i>=1;i--)
	{
		if(has[a[i]]==0)
		{
			has[a[i]]=1;
			lenb++;
			b[lenb]=a[i];
			atob[a[i]]=lenb;
		}
	}
	
	int temp=0x3f3f3f3f;
	for(i=1;i<=m;i++)
	{
		if(has[i]==0)
		{
			temp=i;
			break;
		}
	}
	
	while(lenb>0 && b[lenb]<temp)
	{
		temp=b[lenb];
		lenb--;
	}
	
	
	
	
	memset(calc,0,sizeof(calc));
	calc[1]=n;
	int maxlen=1;
	for(i=q;i>=1;i--)
	{
		int pos=atob[a[i]];
		
		if(pos>lenb)continue;
		
		
		if(calc[pos]>0)
		{
			calc[pos]--;
			calc[pos+1]++;
			
			maxlen=max(maxlen,pos+1);
		}
		else
		{
			if(pos<=maxlen)
			{
				
			}
			else
			{
				break;
			}
		}
		
		/*
		cout<<i<<"    "<<lenb<<"  i"<<endl;
		for(j=1;j<=10;j++)
		{
			cout<<calc[j]<<' ';
		}
		cout<<endl;
		*/
		
	}
	if(i>=1)
	{
		cout<<"No"<<endl;
		return 0;
	}
	
	
	if(maxlen==lenb+1 && calc[lenb+1]==n)
	{
		cout<<"Yes"<<endl;
	}
	else
	{
		cout<<"No"<<endl;
	}
	
	return 0;
}