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 <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const ll INF = 1LL<<29;
const ll mod = 1e9+7;
#define rep(i,n) for(int (i)=0;(i)<(ll)(n);++(i))
#define repd(i,n,d) for(ll (i)=0;(i)<(ll)(n);(i)+=(d))
#define all(v) (v).begin(), (v).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset((m),(v),sizeof(m))
#define chmin(X,Y) ((X)>(Y)?X=(Y),true:false)
#define chmax(X,Y) ((X)<(Y)?X=(Y),true:false)
#define fst first
#define snd second
#define UNIQUE(x) (x).erase(unique(all(x)),(x).end())
template<class T> ostream &operator<<(ostream &os, const vector<T> &v){int n=v.size();rep(i,n)os<<v[i]<<(i==n-1?"":" ");return os;}

#define N 100010
int n, m, q;
int d[N], f[N], r[N], u[N];
vector<int> e;

int main(){
	mset(r, -1);
	cin>>n>>m>>q;
	rep(i, q) cin>>d[i];
	rep(i, q) d[i]--;
	reverse(d, d+q);
	rep(i, q){
		int x = d[i];
		if(r[x]==-1){
			r[x] = e.size();
			e.push_back(x);
			f[0]++;
		} else {
			int y = r[x];
			int lb = 0, ub = n-1; // (lb, ub]
			while(ub-lb>1){
				int md = (lb+ub+1)/2;
				(f[md]>y?lb:ub)=md;
			}
			if(f[ub]==y) f[ub]++;
		}
	}
	//rep(i, n) cerr<<f[i]<<" "; cerr<<endl;
	//rep(i, e.size()) cerr<<e[i]<<" "; cerr<<endl;
	rep(i, f[n-1]) u[e[i]] = 1;
	for(int i = 0, j = f[n-1]; i < m && j < f[0]; i++){
		if(u[i]==0){
			if(e[j]!=i){
				cout<<"No"<<endl;
				return 0;
			} else j++;
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}
