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 <vector>
#include <string>
#include <algorithm>
#include <map>

#define rep(i,n) for(int i=0; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

const int INF = 0x3fffffff;


int main(){
	int N, M, Q;
	bool ans=false;

	cin >> N >> M >> Q;
	vector<int> s;
	vector<int> a(Q);
	rep(i,Q){
		cin >> a[i];
		a[i]--;
	}
	reverse( all(a) );

	map<int,int> mp;
	int max_num=0;
	int cnt[100005]={};
	rep(i,Q){
		if( !mp.count(a[i]) ){
			mp[ a[i] ] = s.size();
			s.push_back(a[i]);
		}
		int num = mp[ a[i] ];
		if( num == 0 || cnt[num-1] > cnt[num] ){
			cnt[num]++;
		}
	}

	map<int,int> m;
	max_num = 0;
	rep(i, s.size()){
		m[ s[i] ] = i;
		if( cnt[i] >= N || s[i] == max_num ){
			ans = true;
		}else{
			ans = false;
			break;
		}
		while( m.count(max_num) ) max_num++;
	}

	cout << (ans?"Yes":"No") << endl;

	return 0;
}
