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;

const int Nmax = 1e5 + 5;

int N, M, Q, i, p, j;
vector<int> v[Nmax];
int a[Nmax], last[Nmax];
bool used[Nmax];

int main()
{
  //  freopen("input", "r", stdin);
    cin.sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> M >> Q;
    for(i=1; i<=Q; ++i)
    {
        cin >> a[i];
        v[a[i]].push_back(i);
    }

    vector<int> ans;

    for(i=Q; i; --i)
        if(!used[a[i]]) ans.push_back(a[i]), used[a[i]] = 1;

    for(i=1; i<=M; ++i)
        if(!used[i]) ans.push_back(i);

    for(i=1; i<=N; ++i) last[i] = Q;

    memset(used, 0, sizeof(used));

    bool bad = 0;
    for(auto it : ans)
    {
        p = v[it].size() - 1;
        for(j=1; j<=N && !bad; ++j)
        {
            while(p>=0 && v[it][p] > last[j]) --p;
            if(p>=0) last[j] = v[it][p], p--;
                else bad = 1;
        }

        if(bad) break;
        used[it] = 1;
    }

    vector<int> ans2;
    for(i=1; i<=M; ++i)
        if(!used[i]) ans2.push_back(i);

    for(i=1; i<=ans2.size(); ++i)
        if(ans[ans.size()-i] != ans2[ans2.size()-i])
        {
            cout << "No\n";
            return 0;
        }

    cout << "Yes\n";
    return 0;
}
