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 <vector>
#include <algorithm>

using namespace std;

int N, M, Q;
int a[100005];
int used[100005];
int pro[100005];

int res[100005];

int main(void)
{
	cin >> N >> M >> Q;
	for(int i = 0; i < Q; i++) cin >> a[i];
	reverse(a, a+Q);
	
	for(int i = 1; i <= M; i++) used[i] = -1;
	for(int i = 0; i < N; i++) pro[i] = -1;
	
	int id = 0;
	for(int i = 0; i < Q; i++){
		if(used[a[i]] == -1){
			used[a[i]] = id++;
			pro[N-1]++;
		}
		else{
			int p = lower_bound(pro, pro+N, used[a[i]]) - pro;
			if(p > 0 && pro[p-1] == used[a[i]]-1){
				pro[p-1]++;
			}
		}
	}
	
	for(int i = 1; i <= M; i++){
		if(used[i] == -1) used[i] = id++;
	}
	for(int i = 1; i <= M; i++) res[used[i]] = i;
	
	vector<int> vec, ans;
	for(int i = 0; i <= pro[0]; i++) vec.push_back(res[i]);
	for(int i = 1; i <= M; i++) vec.push_back(i);
	
	for(int i = 1; i <= M; i++) used[i] = 0;
	for(int i = 0; i < vec.size(); i++){
		if(used[vec[i]] == 0){
			used[vec[i]] = 1;
			ans.push_back(vec[i]);
		}
	}
	
	for(int i = 0; i < ans.size(); i++){
		if(res[i] != ans[i]){
			cout << "No" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;
	
	return 0;
}