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 <iomanip>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    cin.tie(0);
    int n, m, q;
    cin >> n >> m >> q;

    vector<int> as(q);
    for (auto &a: as) {
        cin >> a;
        a--;
    }

    map<int, vector<int>> poss;

    for (int i = 0; i < q; i++) {
        poss[as[i]].push_back(i);
    }

    vector<int> height(n, q);

    set<int> rest;
    for (int i = 0; i < m; i++) {
        rest.insert(i);
    }

    for (int i = q - 1; i >= 0; i--) {
        int cur = as[i];
        if (!rest.count(cur))
            continue;

        auto it = poss[cur].rbegin();
        auto end = poss[cur].rend();
        int partial = -1;

        for (int j = 0; j < height.size(); j++) {
            while (it != end && *it >= height[j])
                it++;

            if (it == end) {
                partial = j;
                break;
            }

            height[j] = *it++;
        }

        if (partial != -1) {
            if (cur != *rest.begin()) {
                cout << "No" << endl;
                return 0;
            }

            while (partial < height.size()) {
                if (height[partial] < 0)
                    break;
                height[partial++] = -1;
            }
        }

        // for (int j = 0; j < height.size(); j++)
        //     cout << i << ", " << j << ": " << height[j] << endl;
        // cout << endl;

        rest.erase(cur);
    }

    cout << "Yes" << endl;

    return 0;
}