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
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<climits>
#include<cmath>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<numeric>
#include<functional>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_set>
#include<random>
#include<array>
#include<cassert>
using namespace std;
#define INF (1<<29)
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(v) v.begin(),v.end()
#define uniq(v) v.erase(unique(all(v)),v.end())

int n, m, q;
int a[100000];

int cnt[100001];
int pos[100000];
bool s[100000];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	vector<int> ord;
	cin >> n >> m >> q;
	rep(i, q) cin >> a[i], a[i]--;

	for (int i = q - 1; i >= 0; i--) {
		if (s[a[i]])continue;
		ord.push_back(a[i]);
		s[a[i]] = true;
	}
	rep(i, m) {
		if (!s[i])ord.push_back(i);
	}

	rep(i, m)pos[ord[i]] = i;
	cnt[0] = n;

	for (int i = q - 1; i >= 0;i--){
		int p = pos[a[i]];
		if (cnt[p] > 0) {
			cnt[p]--;
			cnt[p + 1]++;
		}
	}
	rep(i,m) {
		int p = pos[i];
		cnt[p + 1] += cnt[p];
		cnt[p] = 0;
	}
	cout << (cnt[m] == n ? "Yes" : "No") << endl;


	return 0;
}