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>
#define XX first
#define YY second
#define pb emplace_back
#define FOR(i,a,b) for(int (i)=(a);i<(b);++(i))
#define EFOR(i,a,b) for(int (i)=(a);i<=(b);++(i))
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define REP rep
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define all(X) (X).begin(),(X).end()
#define eb emplace_back

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef ll LL;
typedef pii PII;
typedef pll PLL;
const ll MOD=1e9+7;

#define rall(X) (X).rbegin(),(X).rend()
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rreps(X,S,Y) for (int (X) = (Y)-1;(X) >= (S);--(X))
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}

int N, M, Q;
vector<int> as;
vector<int> vs[114514];
vector<int> needs[114514];

signed main(){
  ios_base::sync_with_stdio(false);
  cout<<fixed<<setprecision(0);
  cin >> N >> M >> Q;
  rep(i, Q) {
    int q;
    cin >> q;
    as.eb(q);
  }

  reps(i, 1, N) needs[0].eb(i);

  set<int> has;
  rrep(i, Q) {
    int t = as[i];
    if (needs[t].empty()) {
      if (has.count(t)) continue;
      has.insert(t);
      vs[0].eb(t);
      needs[t].swap(needs[0]);
    } else {
      int idx = needs[t].back(); needs[t].pop_back();
      vs[idx].eb(t);
      assert(vs[0].size() >= vs[idx].size());
      if (vs[0].size() == vs[idx].size()) needs[0].eb(idx);
      else needs[vs[0][vs[idx].size()]].eb(idx);
    }
  }

  int m = vs[0].size();
  int incr = 0;
  set<int> st;
  rep(i, M) st.insert(i+1);
  rep(i, m) {
    int v = vs[0][i];
    if (st.count(v)) {
      if (v != *st.begin()) incr = i+1;
      st.erase(v);
    }
  }

  reps(i, 1, N) {
    if (incr > vs[i].size()) {
      cout << "No\n";
      return 0;
    }
  }
  cout << "Yes\n";
}
