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 <vector>
#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
using namespace std;

const string NO = "No";
const string YES = "Yes";

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    int q;
    cin >> q;

    vector<int> a(q);
    for (int i = 0; i < q; ++i) {
        cin >> a[i];
    }

    reverse(a.begin(), a.end());

    vector<int> b(n - 1);
    for (int i = 0; i + 1 < n; ++i) {
        b[i] = 0;
    }

    set<int> values;
    for (int x = 1; x <= m; ++x) {
        values.insert(x);
    }

    vector<int> minValues(m);
    for (int i = 0; i < m; ++i) {
        minValues[i] = 1;
    }

    map<int, int> mp;
    int last = 1;

    vector<int> revValues(m + 1);

    for (int x : a) {
        if (mp.find(x) == mp.end()) {
            values.erase(values.find(x));
            if (last < m) {
                minValues[last] = (*values.begin());
            }
            revValues[last] = x;
            mp[x] = last++;
            continue;
        }

        int xx = mp[x] - 1;

        int bl = 0;
        int br = n - 2;
        int ba = -1;
        while (bl <= br) {
            int bm = (bl + br) >> 1;

            if (b[bm] == xx) {
                ba = bm;
                br = bm - 1;
            } else if (b[bm] > xx) {
                bl = bm + 1;
            } else {
                br = bm - 1;
            }
        }

        if (ba == -1) {
            continue;
        } else {
            ++b[ba];
        }
    }

    vector<bool> good(last);

    good[last - 1] = true;
    for (int i = last - 2; i >= 0; --i) {
        good[i] = good[i + 1];
        if (revValues[i + 1] != minValues[i]) {
            good[i] = false;
        }
    }

    for (int i = 0; i + 1 < n; ++i) {
        if (!good[b[i]]) {
            cout << NO << endl;
            return 0;
        }
    }

    cout << YES << endl;

    return 0;
}