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 <fstream>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <queue>

#define mp make_pair
#define pb push_back


typedef long long ll;
typedef long double ld;

using namespace std;
const int MX = 120000;

int n, m, q;
int a[MX];
vector<int> vv;
int us[MX];
int cnt[MX];

int main() {
	cin >> n >> m;
	cin >> q;
	for (int i = 0; i < q; ++i)
		cin >> a[i], --a[i];
	for (int i = 0; i < m; ++i)
		us[i] = -1;
	for (int i = q - 1; i >= 0; --i) {
		if (us[a[i]] == -1) {
			us[a[i]] = vv.size();
			vv.push_back(a[i]);
		}
	}
	for (int i = 0; i < m; ++i)
		if (us[i] == -1)
			us[i] = vv.size(), vv.push_back(i);
	int nd = vv.size() - 1;
	for (int i = (int)vv.size() - 1; i >= 1; --i) {
		if (vv[i] > vv[i - 1])
			--nd;
		else
			break;
	}
	cnt[0] = n;
	for (int i = q - 1; i >= 0; --i) {
		int x = us[a[i]];
		if (cnt[x] != 0)
			--cnt[x], ++cnt[x + 1];
	}
	for (int i = 0; i < nd; ++i)
		if (cnt[i] != 0) {
			cout << "No\n";
			return 0;
		}
	cout << "Yes\n";

	return 0;
}


