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;
typedef long long li;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    li n, m;
    cin >> n >> m;
    li q;
    cin >> q;
    vector<li> qs(q);
    for (int i = 0; i < q; ++i) {
        cin >> qs[i];
        qs[i]--;
    }

    reverse(qs.begin(), qs.end());
    bool ok = true;

    map<li, li> positions;
    map<li, li> length_count;
    length_count[0] = n;
    for (int i = 0; i < q; ++i) {
        bool first_time = positions.count(qs[i]) == 0;
        if (first_time) {
            const li next_index = positions.size();
            positions[qs[i]] = next_index;
        }

        const li next_index = positions[qs[i]];
        if (length_count[next_index] > 0) {
            length_count[next_index]--;
            length_count[next_index + 1]++;
        } else if (first_time) {
            ok = false;
            break;
        }
    }

    if (ok) {
        li shortest = m;
        for (int i = 0; i < m; ++i) {
            if (length_count[i] > 0) {
                shortest = min(shortest, (li)i);
            }
        }
        vector<li> order(positions.size());
        for (auto &position  : positions) {
            order[position.second] = position.first;
        }
        set<li> unseen;
        for (int i = 0; i < m; ++i) {
            unseen.insert(i);
        }
        for (int i = 0; i < positions.size(); ++i) {
            if (i < shortest) {
                unseen.erase(order[i]);
            } else {
                ok &= unseen.lower_bound(order[i]) == unseen.begin();
                unseen.erase(order[i]);
            }
        }
    }

    cout << (ok ? "Yes" : "No") << endl;
    return 0;
}