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
#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <complex>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <bitset>
#include <numeric>
#include <limits>
#include <climits>
#include <cfloat>
#include <functional>
using namespace std;

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

    vector<int> index(m, -1);
    vector<int> cnt(m, 0);
    vector<int> v;
    for(int i=q-1; i>=0; --i){
        if(index[a[i]] == -1){
            index[a[i]] = v.size();
            v.push_back(a[i]);
        }
        int j = index[a[i]];
        if(j == 0 || cnt[v[j]] < cnt[v[j-1]])
            ++ cnt[a[i]];
    }

    unsigned i = 0;
    while(i < v.size() && cnt[v[i]] >= n)
        ++ i;

    int j = 0;
    while(i < v.size()){
        while(j < m && cnt[j] >= n)
            ++ j;
        if(v[i] != j){
            cout << "No" << endl;
            return 0;
        }
        ++ i;
        ++ j;
    }

    cout << "Yes" << endl;
    return 0;
}