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 <vector>
#include <algorithm>
using namespace std;
#define all(x) (x).begin(), (x).end()

int n, m, q;
int a[500000];
vector<int> line;
int pos[500000];
int cnt[500000];

signed main()
{
    cin >> n >> m >> q;
    for (int i = 0; i < q; i++) cin >> a[i];
    fill(pos, pos + m + 1, -1);
    for (int i = q - 1; i >= 0; i--)
    {
        int x = a[i];
        if (pos[x] == -1)
        {
            pos[x] = line.size();
            cnt[x]++;
            line.push_back(x);
        }
        else
        {
            if (pos[x] == 0)
            {
                cnt[x]++;
                continue;
            }
            else
            {
                if (cnt[line[pos[x] - 1]] > cnt[x])
                {
                    cnt[x]++;
                }
            }
        }
    }
    int t = 1;
    for (int i = 0; i < line.size(); i++)
    {
        if (cnt[line[i]] < n)
        {
            t = 0;
        }
    }
    if (t) cout << "Yes";
    else
    {
        vector<int> g, f;
        for (int i = 0; i < line.size(); i++)
        {
            g.push_back(line[i]);
            if (cnt[line[i]] >= n) f.push_back(line[i]);
        }
        for (int i = 1; i <= m; i++)
        {
            if (cnt[i] == 0) g.push_back(i);
            if (cnt[i] < n) f.push_back(i);
        }
        //for (auto e : g) cout << e << " "; cout << endl;
        //for (auto e : f) cout << e << " "; cout << endl;
        if (g == f)
        {
            cout << "Yes";
        }
        else
        {
            cout << "No";
        }
    }
}
