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
// In the name of God

#include <iostream>
#include <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>

using namespace std;

#define rep(i, n) for (int i = 0, _n = (int)(n); i < _n; ++i)
const int N = (int) 1e6 + 6, mod = (int) 0;
vector<int> ord;
int markn[N], a[N], cnt[N], pl[N];
int32_t main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int n, m, q;
    cin >> n >> m >> q;
    for (int i = 0; i < q; ++i) {
        cin >> a[i];
        a[i]--;
    }
    cnt[0] = n;
    for (int i = q - 1; i >= 0; --i) {
        if (!markn[a[i]]++) {
            ord.push_back(a[i]);
        }
    }
    for (int i = 0; i < m; ++i)
        if (!markn[i]++) {
            ord.push_back(i);
        }
    for (int i = 0; i < m; ++i)
        pl[ord[i]] = i;
    for (int i = q - 1; i >= 0; --i) {
        if (cnt[pl[a[i]]] == 0) continue;
        cnt[pl[a[i]]]--;
        cnt[pl[a[i]] + 1]++;
    }
    int sr = 0;
    for (int i = m - 2; i >= 0; --i) {
        if (ord[i + 1] < ord[i]) {
            sr = i + 1;
            break;
        }
    }
    for (int i = 0; i < sr; ++i) {
        if (cnt[i] > 0) {
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";


}



