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 <stack>
#include <set>
using namespace std;

const int MAXN = 100005;

int N, M, Q;
int a[MAXN];
stack <int> occ[MAXN], pos[MAXN];
set <int> S;

void load() {
	scanf("%d%d%d", &N, &M, &Q);
	for (int i = 0; i < Q; i++)
		scanf("%d", a + i);
}

bool solve() {
	for (int i = 0; i < Q; i++)
		occ[a[i]].push(i);
	
	for (int i = 1; i <= M; i++)
		S.insert(i);
	
	for (int i = 0; i < N; i++)
		pos[i].push(Q);
	
	bool ok = true;
	for (int i = Q - 1; i >= 0; i--) {
		if (!S.count(a[i])) continue;
		for (int j = 0; j < N; j++) {
			while (!occ[a[i]].empty() && occ[a[i]].top() >= pos[j].top())
				occ[a[i]].pop();
			if (occ[a[i]].empty()) {
				ok = false;
				break;
			}
			pos[j].push(occ[a[i]].top());
			occ[a[i]].pop();
		}
		if (!ok && a[i] != *S.begin()) 
			return false;
		S.erase(a[i]);
	}
	
	return true;
}

int main() {
	load();
	puts(solve() ? "Yes" : "No");
	return 0;
}