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 <cstdio>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <cmath>
#include <string>
#include <map>
#include <unordered_map>
#include <queue>
#include <climits>

using namespace std;

typedef pair<int, int> PII;

const int M = 100005;

int n, m, q;

int a[M];
bool used[M];
int cnt[M], ord[M];
vector<int> seq;

void add(int x) {
  if (!used[x]) {
    int f = seq.size();
    used[x] = true;
    ord[x] = f;
    if (cnt[f] > 0) {
      cnt[f] -= 1;
      cnt[f + 1] += 1;
    }
    seq.push_back(x);
  } else {
    int k = ord[x];
    if (cnt[k] > 0) {
      cnt[k] -= 1;
      cnt[k + 1] += 1;
    }
  }
}

int main() {
  cin >> n >> m >> q;
  for (int i = 0; i < q; i += 1) {
    cin >> a[i];
    a[i] -= 1;
  }
  reverse(a, a + q);
  cnt[0] = n;
  for (int i = 0; i < q; i += 1) add(a[i]);
  for (int i = 0; i < m; i += 1) if (!used[i]) seq.push_back(i);
  int sh = 0;
  while (!cnt[sh]) sh += 1;
  auto alt = seq;
  sort(alt.begin() + sh, alt.end());
  if (alt == seq) {
    cout << "Yes\n";
  } else {
    cout << "No\n";
  }
}
