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 <string>
#include <vector>
#include <set>
#include <algorithm>

using namespace std;

int main(){
	ios_base::sync_with_stdio(false);
	int n, m, q;
	cin >> n >> m >> q;
	vector<int> a(q);
	for(int i = 0; i < q; ++i){
		cin >> a[i];
		--a[i];
	}
	vector<int> occur(m);
	vector<int> order;
	for(int i = q - 1; i >= 0; --i){
		if(occur[a[i]] == 0){ order.push_back(a[i]); }
		++occur[a[i]];
	}
	const int k = order.size();
	bool is_ordered = true;
	for(int i = 0; i < k; ++i){
		if(order[i] != i){ is_ordered = false; }
	}
	if(is_ordered){
		cout << "Yes" << endl;
		return 0;
	}
	vector<int> lookup(m);
	for(int i = 0; i < k; ++i){ lookup[order[i]] = i; }
	vector<int> reach(k + 1);
	reach[0] = n;
	for(int i = q - 1; i >= 0; --i){
		const int t = lookup[a[i]];
		if(reach[t + 1] < reach[t]){ ++reach[t + 1]; }
	}
	set<int> s;
	for(int i = 0; i < m; ++i){ s.insert(i); }
	bool possible = true;
	for(int i = 0; possible && i < k; ++i){
		if(reach[i + 1] < n && *(s.begin()) != order[i]){
			possible = false;
		}
		s.erase(order[i]);
	}
	cout << (possible ? "Yes" : "No") << endl;
	return 0;
}

