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>
using namespace std;

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, m, q;
  cin >> n >> m >> q;
  vector<int> a(q), c(m);
  for (auto&& e : a) {
    cin >> e;
    --e;
    ++c[e];
  }
  reverse(begin(a), end(a));
  set<int> se;
  vector<int> v;
  for (int e : a) {
    if (se.count(e)) {
      continue;
    }
    se.insert(e);
    v.push_back(e);
  }
  for (int i = 0; i < m; ++i) {
    if (se.count(i)) {
      continue;
    }
    se.insert(i);
    v.push_back(i);
  }
  while (v.size() >= 2 and v[v.size() - 2] < v.back()) {
    v.pop_back();
  }
  v.pop_back();
  if (any_of(begin(v), end(v), [&](int e) { return c[e] < n; })) {
    cout << "No\n";
    exit(0);
  }
  int k = v.size();
  map<int, int> mp;
  for (int i = 0; i < k; ++i) {
    mp[v[i]] = i;
  }
  vector<int> t(k);
  for (int e : a) {
    if (not mp.count(e)) {
      continue;
    }
    int i = mp[e];
    if (t[i] == n) {
      continue;
    }
    if (i == 0 or t[i - 1] > t[i]) {
      ++t[i];
    }
  }
  if (all_of(begin(t), end(t), [&](int e) { return e == n; })) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
}
