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 <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n, m, q, a[N], vec[N], cnt[N], pos[N];
set<int> st;
int main() {
	scanf("%d%d%d", &n, &m, &q);
	for (int i = 1; i <= q; i++) {
		scanf("%d", &a[i]);
	}
	for (int i = 1; i <= m; i++) st.insert(i);
	int tmp = 0;
	for (int i = q; i >= 1; i--) {
		if (st.count(a[i])) {
			vec[++tmp] = a[i];
			st.erase(a[i]);
		}
	}
	while (st.size()) {
		vec[++tmp] = *st.begin();
		st.erase(st.begin());
	}
	for (int i = 1; i <= m; i++) pos[vec[i]] = i;
	for (int i = q; i >= 1; i--) {
		int p = pos[a[i]];
		cnt[p] = min(cnt[p] + 1, n);
		if (p > 1 && cnt[p - 1] < cnt[p]) {
			if (!cnt[p - 1]) {
				puts("No");
				return 0;
			}
			cnt[p] = cnt[p - 1];
		}
	}
	int st;
	for (st = m; st > 1 && vec[st - 1] < vec[st]; st--);
	for (int i = 1; i < st; i++) {
		if (cnt[i] < n) {
			puts("No");
			return 0;
		}
	}
	puts("Yes");
	return 0;
}