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;
typedef long long LL;
typedef pair<int,int> PII;

int main() {
	int N,M, Q;
	cin >> N >> M >> Q;
	
	map<int, int> order;
	vector<int> input;
	set<int> used;
	vector<int> pos(M+1, 0);
	
	pos[0] = N;
	
	vector<int> nums;
	for(int i=0; i<Q; i++) {
		int p;
		cin >> p;
		input.push_back(p);
	}
	reverse(input.begin(), input.end());
	for(int i=0; i<input.size(); i++) {
		int p = input[i];
		int cnt = used.size();
		used.insert(p);
		if((int)used.size() != cnt) {
			order[p] = cnt+1;
			nums.push_back(p);
		}
		int cur = order[p];
		if(pos[cur-1] > 0) {
			pos[cur-1]--;
			pos[cur]++;
		}
	}
	/*
	for(int i=0; i<3; i++) {
		cout <<" " << i << " " << pos[i] << endl;
	}
	for(int i=0; i<nums.size(); i++)
		cout << nums[i] << endl;
	*/
	int last = nums.size()-1;
	bool ok = true;
	for(int val = nums.back(); val>=1; val--) {
		
		if(order[val] == 0) ok = false;
		if(last>0 && val == nums[last-1]) {
			last--;
		}
	}
	if(!ok) {
		last = nums.size();
	}
	
	bool judge = true;
	for(int i=0; i<last; i++)
		if(pos[i] > 0)
			judge = false;
	if(judge) cout << "Yes" << endl;
	else cout << "No" << endl;	
	return 0;
}
