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;
typedef long long ll;

const ll p = 1e9+7;
const int MAXN = 1e5+5;

int n, m, q;
int a[MAXN], take[MAXN];
vector<int> pl[MAXN];

map<int, int> s;

int main() {
	cin >> n >> m >> q;
	for (int i = 0; i < q; ++i) {
		cin >> a[i]; a[i]--;
		pl[a[i]].push_back(i);
	}
	for (int i = 0; i < m; ++i)
		s[i] = 1;

	for (int i = 0; i < n; ++i)
		take[i] = q;
	for (int i = q-1; i >= 0; --i) {
		if (s.empty())
			break;
		if (s.count(a[i]) == 0)
			continue;
		int fir = (*s.begin()).first;
		s.erase(a[i]);
		int ct = n-1;
		if (fir == a[i]) {
			for (int j = pl[a[i]].size()-1; j >= 0; --j) {
				if (ct == -1)
					break;
				if (pl[a[i]][j] <= take[ct]) {
					take[ct] = pl[a[i]][j]; ct--;
				}
			}

			for (int j = ct; j >= 0; --j)
				take[ct] = -1;
		} else {
			for (int j = pl[a[i]].size()-1; j >= 0; --j) {
				if (ct == -1)
					break;
				if (pl[a[i]][j] <= take[ct]) {
					take[ct] = pl[a[i]][j]; ct--;
				}
			}

			if (ct >= 0) {
				cout << "No\n"; return 0;
			}
		}
	}

	cout << "Yes\n";
}