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 <deque>
#include <map>
#include <algorithm>
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }


#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl
template<class T1, class T2> ostream& operator << (ostream &s, pair<T1,T2> P)
{ return s << '<' << P.first << ", " << P.second << '>'; }
template<class T> ostream& operator << (ostream &s, vector<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, deque<T> P)
{ for (int i = 0; i < P.size(); ++i) { if (i > 0) { s << " "; } s << P[i]; } return s; }
template<class T> ostream& operator << (ostream &s, vector<vector<T> > P)
{ for (int i = 0; i < P.size(); ++i) { s << endl << P[i]; } return s << endl; }

#define EACH(i, s) for (__typeof__((s).begin()) i = (s).begin(); i != (s).end(); ++i)
template<class T1, class T2> ostream& operator << (ostream &s, map<T1,T2> P)
{ EACH(it, P) { s << "<" << it->first << "->" << it->second << "> "; } return s << endl; }



int N, M, Q;
deque<int> a;

bool solve() {
    vector<int> b(M);
    for (int i = 0; i < M; ++i) b[i] = i;
    map<int,int> id;
    int iter = 0;
    for (int i = 0; i < a.size(); ++i) {
        if (id.count(a[i])) continue;
        id[a[i]] = iter++;
    }
    for (int i = 0; i < b.size(); ++i) {
        if (id.count(b[i])) continue;
        id[b[i]] = iter++;
    }
    for (int i = 0; i < a.size(); ++i) a[i] = id[a[i]];
    for (int i = 0; i < b.size(); ++i) b[i] = id[b[i]];

    vector<int> con(id.size(), 0);
    for (int i = 0; i < a.size(); ++i) {
        if (a[i] == 0) con[a[i]]++;
        else if (con[a[i]] + 1 <= con[a[i]-1]) con[a[i]]++;
    }
    for (int i = 0; i < b.size(); ++i) {
        if (b[i] == 0) chmax(con[b[i]], N);
        else chmax(con[b[i]], con[b[i]-1]);
    }
    bool ok = true;
    for (int i = 0; i < con.size(); ++i) {
        if (con[i] < N) ok = false;
    }

    //COUT(a); COUT(id); COUT(a); COUT(con);
    
    return ok;
}

int main() {
    while (cin >> N >> M >> Q) {
        a.resize(Q);
        for (int i = 0; i < Q; ++i) cin >> a[i], --a[i];
        reverse(a.begin(), a.end());
        if (a.empty() || solve()) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
}
