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 <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int N = 100100;
int n, m, q;
int a[N];
int pos[N];
int len[N];
int cnt[N];
int b[N];

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);


	scanf("%d%d%d", &n, &m, &q);
	for (int i = 0; i < q; i++)
		scanf("%d", &a[i]);
	reverse(a, a + q);
	for (int i = 1; i <= m; i++)
		pos[i] = -1;
	for (int i = 0; i < q; i++)
	{
		int x = a[i];
		if (pos[x] == -1)
		{
			pos[x] = len[0];
			b[len[0]] = x;
			len[0]++;
			cnt[x] = 1;
		}
		else
		{
			int y = cnt[x];
			if (y >= n) continue;
			if (len[y] < pos[x]) continue;
			len[y]++;
			cnt[x]++;
		}
	}
	/*
	for (int i = 0; i < n; i++)
		printf("%d ", len[i]);
	printf("\n");
	for (int i = 0; i < len[0]; i++)
		printf("%d ", b[i]);
	printf("\n");
	*/
	int minUnused = m + 1;
	for (int x = 1; x <= m; x++)
		if (pos[x] == -1)
		{
			minUnused = x;
			break;
		}
	int L = len[0];
	while(L > 0 && b[L - 1] < minUnused)
	{
		L--;
		minUnused = b[L];
	}
	bool ok = 1;
	for (int i = 0; i < n; i++)
		ok &= len[i] >= L;
	if (ok)
		printf("Yes\n");
	else
		printf("No\n");

	return 0;
}