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
//#pragma comment(linker, "/STACK:102400000,102400000") 
#pragma warning(disable:4996)
#include <fstream>
#include <iostream>
#include <functional>
#include <algorithm>
#include <cstring>
#include <vector>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <deque>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;

typedef long long ll;
#define eps 1e-10
#define LL_INF 0x3fffffffffffffff
#define INF 0x3f3f3f3f
#define mem(a, b) memset(a, b, sizeof(a))
#define pper(i,n,m) for(int i = n;i >= m; i--)
#define repp(i, n, m) for (int i = n; i <= m; i++)
#define rep(i, n, m) for (int i = n; i < m; i++)
#define sa(n) scanf("%d", &(n))
#define mp make_pair
#define ff first
#define ss second
#define pb push_back

const int maxn = 5005;
const ll mod = 1e9 + 7;
const double PI = acos(-1.0);
ll po(ll a, ll b, ll mod) { ll res = 1; a %= mod; for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; }
ll gcd(ll a, ll b) { if (a == 0) { return b; } else { return gcd(b%a, a); } }

int N, M, Q;
int A[101010];
int did[101010];
int ind[101010];
vector<int> V;
int num[101010];

void solve() {
	int i, j, k, l, r, x, y; string s;

	cin >> N >> M >> Q;
	rep(i, 0, Q)
	{
		cin >> A[i];
	}
	for (i = Q - 1; i >= 0; i--)
	{
		did[A[i]]++;
		if (did[A[i]] == 1)
		{
			ind[A[i]] = V.size();
			V.push_back(A[i]);
		}
	}
	rep(i, 0, M)
	{
		if (did[i + 1] == 0)
		{
			ind[i + 1] = V.size();
			V.push_back(i + 1);
		}
	}

	while (V.size() >= 2 && V[V.size() - 2]<V[V.size() - 1]) V.pop_back();
	V.pop_back();

	num[0] = N;
	for (i = Q - 1; i >= 0; i--) 
	{
		if (num[ind[A[i]]])
		{
			num[ind[A[i]]]--;
			num[ind[A[i]] + 1]++;
		}
	}

	rep(i, 0, V.size())
	{
		if (num[i])
		{
			printf("No\n");
			return;
		}
	}
	printf("Yes\n");
}

int main()
{
	solve();
	return 0;
}


