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 <cmath>
#include <cstring>
#include <ctime>
#include <climits>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <sstream>
#include <typeinfo>
#include <fstream>

#define DIV 1000000007

using namespace std;

long long N, M, Q;
long long a[100005];
long long b[100005];
long long idx[100005];

vector<long long> ans;
set<long long> done;

int main(){
	cin >> N >> M >> Q;
	for(int i = 0; i < Q; i++){
		cin >> a[i];
		a[i]--;
	}

	//create correct
	for(int i = Q - 1; i >= 0; i--){
		if(done.count(a[i]) == 0){
			ans.push_back(a[i]);
			done.insert(a[i]);
		}
	}
	for(int i = 0; i < M; i++){
		if(done.count(i) == 0){
			ans.push_back(i);
			done.insert(i);
		}
	}

	for(int i = 0; i < ans.size(); i++){
		idx[ans[i]] = i;
	}
	b[0] = N;
	
	for(int i = Q - 1; i>=0;i--){
		if(b[idx[a[i]]] > 0){
			b[idx[a[i]]]--;
			b[idx[a[i]]+1]++;
		}
	}

	long long mini = M;
	for(long long i = 0; i < M; i++){
		//cout << "b[" << i << "]=" << b[i] << endl;
		if(b[i]>0){
			mini = min(mini, i);
		}
	}

	done.clear();
	for(int i = 0; i < mini; i++){
		done.insert(ans[i]);
	}
	long long idx = mini;

	for(int i = 0; i < M; i++){
		if(done.count(i) == 0){
			//cout << "i=" << i << " idx = " << idx << endl;
			if(ans[idx] != i){
				cout << "No" << endl;
				return 0;
			}
			idx++;
		}
	}
	cout << "Yes" << endl;


	//for(long long i = mini; i < M; i++){
	//	if(ans[i] != i){
	//		cout << "No" << endl;
	//		return 0;
	//	}
	//}
	//cout << "Yes" << endl;
}
