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 <cstdlib>
#include <cstring>
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <set>
#include <cmath>
#include <map>
using namespace std;
#define MOD 1000000007
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;

int N, M, Q;
int A[101010];
int cnt[101010];
set<int> poss[101010];

int main()
{
	scanf("%d%d%d", &N, &M, &Q);
	for (int i = 0; i < Q; ++i) {
		scanf("%d", A + i);
		--A[i];
	}
	for (int i = 0; i < Q; ++i) {
		++cnt[A[i]];
		poss[A[i]].insert(i);
	}
	set<int> rem;
	for (int i = 0; i < M; ++i) rem.insert(i);
	vector<int> req;
	for (int i = Q - 1; i >= 0; --i) {
		if (rem.count(A[i]) == 0) continue;
		if (0 && rem.size() > 0 && *(rem.begin()) == A[i]) {
			// you can ignore it
		} else {
			req.push_back(A[i]);
		}
		rem.erase(A[i]);
	}
	for (int i : rem) req.push_back(i);
	vector<bool> ok(req.size(), true);
	for (int i = 0; i < req.size() - 1; ++i) if (req[i] > req[i + 1]) ok[i] = false;
	for (int i = req.size() - 2; i >= 0; --i) ok[i] = ok[i] && ok[i + 1];

	for (int i = 0; i < N; ++i) {
		int lp = Q;
		int fg = -1;
		for (int _j = 0; _j < req.size(); _j++) {
			int j = req[_j];
			int nxtp = -1;
			auto pt = poss[j].upper_bound(lp);
		//	printf("%d ", lp);
			if (pt == poss[j].begin()) {
				if (ok[_j]) break;
				else {
					puts("No");
					return 0;
				}
			} else {
				nxtp = *(--pt);
				lp = nxtp;
				poss[j].erase(lp);
			}
		}
	//	puts("");
	}
	puts("Yes");

	return 0;
}
