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 <cstring>
#include <vector>
#include <algorithm>
using namespace std;


int main()
{
	int n, m;
	int q;
	vector<int> a;

	scanf("%d%d%d", &n, &m, &q);
	a.resize(q);
	for(int i = 0; i < q; ++i)
		scanf("%d", &a[i]);

	vector<int> sample;
	vector<bool> exist(m);

	for(int i = q - 1; i >= 0; --i) {
		int x = a[i];
		if(exist[x]) continue;
		exist[x] = true;
		sample.push_back(x);
	}
	for(int x = 1; x <= m; ++x) {
		if(exist[x]) continue;
		sample.push_back(x);
	}

	vector<int> pos(m + 1, -1);

	for(int i = 0; i < m; ++i)
		pos[sample[i]] = i;

	vector<int> count(m + 1);
	count[0] = n;

	for(int i = q - 1; i >= 0; --i) {

		int x = a[i];
		int p = pos[x];

		if(count[p] > 0) {
			count[p] -= 1;
			count[p + 1] += 1;
		}
	}

	int border = m;

	for(int p = m - 1; p >= 1; --p) {
		if(sample[p - 1] > sample[p]) break;
		border = p;
	}
	if(border == 1) border -= 1;

	bool ok = true;

	for(int i = 0; i < border - 1; ++i) {
		if(count[i] > 0)
			ok = false;
	}

	/*for(int i = 0; i <= m; ++i) {
		printf("count[%d] = %d\n", i, count[i]);
	}
	for(int i = 0; i < m; ++i) {
		printf("sample[%d] = %d\n", i, sample[i]);
	}
	printf("border: %d\n", border);*/

	printf("%s\n", ok ? "Yes" : "No");

	return 0;
}
