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 {{{
#include <iostream>
#include <cassert>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <bitset>
#include <vector>
#include <complex>
#include <algorithm>
using namespace std;
// }}}
// #define {{{
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
typedef vector<int> vi;
#define de(x) cout << #x << "=" << x << endl
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define per(i,a,b) for(int i=(b)-1;i>=(a);--i)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
// }}}

const int N = 1e5 + 10;
int n , m , Q , a[N] , v[N] , _;
int vis[N] , cnt[N] , pos[N];

int main(){
  scanf("%d%d%d",&n,&m,&Q);
  rep(i,0,Q) scanf("%d",a + i);
  reverse(a,a+Q);
  rep(i,0,Q) {
    if(!vis[a[i]])
      vis[a[i]]=1,v[++_]=a[i];
  }
  rep(i,1,m+1) if(!vis[i])
    v[++_]=i;
  rep(i,1,m+1) pos[v[i]]=i;
  cnt[0]=n;
  int Max=0;
  rep(i,0,Q) {
    int t=pos[a[i]];
    if(cnt[t-1]) {
      cnt[t-1]--;
      cnt[t]++;
      Max = max(Max , t);
    } else if(Max < t) {
      puts("No");
      return 0;
    }
  }
  int Min=0;
  rep(i,0,m+1) if(cnt[i]) {
    Min=i;
    break;
  }
  int inc=true;
  rep(i,Min+2,m+1) inc&=v[i]>v[i-1];
  puts(inc ? "Yes" : "No");
  return 0;
}
