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 <bits/stdc++.h>
using namespace std;

#define FOR(i,l,r) for(int i = (l);i < (r);i++)
#define ALL(x) (x).begin(),(x).end()
template<typename T> void chmax(T& a,const T& b){if(a < b) a = b;}
template<typename T> void chmin(T& a,const T& b){if(b < a) a = b;}
typedef long long ll;

int N,M,Q;

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> N >> M >> Q;
	vector<int> A(Q);
	FOR(i,0,Q){
		cin >> A [i];
	}

	set<int> st;
	vector<int> B,pos(M + 1,-1);
	for(int i = Q - 1;i >= 0;i--){
		if(st.count(A [i]) == 0){
			pos [A [i]] = B.size();
			B.push_back(A [i]);
			st.insert(A [i]);
		}
	}

	vector<int> cnt(B.size() + 1);
	cnt [0] = N;
	for(int i = Q - 1;i >= 0;i--){
		int p = pos [A [i]];
		if(cnt [p]){
			cnt [p]--;
			cnt [p + 1]++;
		}
	}
	
	st.clear();
	vector<int> C;
	for(int i = 0;i < B.size() && cnt [i] == 0;i++){
		C.push_back(B [i]);
		st.insert(B [i]);
	}

	for(int i = 1;i <= M && C.size() < B.size();i++){
		if(st.count(i) == 0){
			C.push_back(i);
		}
	}

	cout << (B == C ? "Yes" : "No") << endl;

	return 0;
}
