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<vector>
#include<algorithm>
#include<set>
#include<map>
#include<unordered_map>
#include<queue>
#include<iomanip>
#include<math.h>
#include<bitset>
#include<cassert>
#include<random>
#include<time.h>
#include<functional>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007LL
#define INF 1000000000LL
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v)  sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

/* --------------------------------------- */

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n, m;
	cin >> n >> m;
	ll q;
	cin >> q;
	vector<ll> a(q);
	REP(i, q) cin >> a[i];
	REP(i, q) a[i]--;
	vector<ll> cnt(m, 0);
	map<ll, ll> dict;
	vector<ll> idx(m, -1);
	ll ite = 0;
	for(ll i = q - 1; i >= 0; i--) {
		if(dict.count(a[i])) {
			ll pos = dict[a[i]];
			if(cnt[pos] == n) continue;
			if(pos != 0 && cnt[pos - 1] <= cnt[pos]) continue;
			cnt[pos]++;
		} else {
			dict[a[i]] = ite;
			idx[ite] = a[i];
			cnt[ite]++;
			ite++;
		}
	}
	vector<ll> not_comp;
	REP(i, m) if(!dict.count(i) || cnt[dict[i]] != n) not_comp.pb(i);
	sort(ALL(not_comp));
	ll tmp = 0;
	REP(i, m) {
		if(cnt[i] == 0 || cnt[i] == n) continue;
		if(idx[i] != not_comp[tmp]) {
			cout << "No" << endl;
			return 0;
		}
		tmp++;
	}
	cout << "Yes" << endl;
	return 0;
}

/* --------------------------------------- */

