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<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<cmath>
#include<climits>
#include<string>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define rep(i,n) for(int i=0;i<((int)(n));i++)
#define reg(i,a,b) for(int i=((int)(a));i<=((int)(b));i++)
#define irep(i,n) for(int i=((int)(n))-1;i>=0;i--)
#define ireg(i,a,b) for(int i=((int)(b));i>=((int)(a));i--)
typedef long long int lli;
typedef pair<int,int> mp;
#define fir first
#define sec second
#define IINF INT_MAX
#define LINF LLONG_MAX
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#define pque(type) priority_queue<type,vector<type>,greater<type> >
#define memst(a,b) memset(a,b,sizeof(a))
#define iter(v,ite) for(auto ite=(v).begin();ite!=(v).end();ite++)
#define mimunum(v,x) distance((v).begin(),lower_bound((v).begin(),(v).end(),x))


lli cn[100005];


int n,m,qn;
int dat[100005];
int gone[100005]={};


vector<int> td;

int main(void){
	scanf("%d%d%d",&n,&m,&qn);
	rep(i,qn)scanf("%d",&dat[i]);
	
	int bp=1;
	irep(i,qn){
		int no=dat[i];
		if(!gone[no]){
			td.push_back(no);
			gone[no]=bp;
			bp++;
		}
	}
	
	reg(i,1,m){
		if(!gone[i])td.push_back(i);
	}
	
	/*
	rep(i,m){
		printf("td .. %d\n",td[i]);
	}
	*/
	
	cn[0]=n;
	
	irep(i,qn){
		int no=dat[i],
			p=gone[no];
		if(!cn[p-1])continue;
		cn[p-1]--;
		cn[p]++;
	}
/*
2 3
4
3 2 3 1
*/
	/*
	rep(i,m+1){
		printf("cn[%d] %d\n",i,cn[i]);
	}
	*/
	
	
	int okp=-1;
	int nma=100005;
	irep(i,m){
		if(nma<td[i]){
			okp=i;
			break;
		}
		nma=td[i];
	}
	//printf("%d %d\n",nma,okp);
	
	reg(i,0,okp){
		if(cn[i]>0){
			printf("No\n");
			return 0;
		}
	}
	
	printf("Yes\n");
	return 0;
}




