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
// ryan in ny / per jorner fills in: derivative code
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<(n);i++)
#define per(i,a,n) for (int i=(n)-1;i>=(a);i--)
typedef long long ll;


int main() {
    ios_base::sync_with_stdio(false);
    int n, m, q;
    cin >> n >> m >> q;
    
    vector<int> v(q + 10);
    rep(i, 0, q) cin >> v[i], v[i]--;
    
    vector<int> waiting(m + 10, 0);
    waiting[0] = n;
    vector<int> pos(m + 10, -1);
    int built = 0;
    vector<int> order(m + 10, -1);
    
    per(i, 0, q) {
        if (pos[v[i]] != -1) {
            if (waiting[pos[v[i]]]) {
                waiting[pos[v[i]]]--;
                waiting[pos[v[i]] + 1]++;
            }
        } else {
            pos[v[i]] = built;
            assert(waiting[built]);
            waiting[built]--;
            waiting[built + 1]++;
            order[built] = v[i];
            built++;
        }
    }
    
    int mn = m;
    rep(i, 0, n) if (pos[i] == -1) {
        mn = i;
        break;
    }
    if (mn != m) order[built] = mn;
    //cout << mn << endl;
    
    //rep(i, 0, m) cout << order[i] << ": " << waiting[i] << endl;
    
    
    
    bool ok = true;
    per(i, 0, m - 1) {
        if (order[i + 1] != -1 && order[i] > order[i + 1]) ok = false;
        if (!ok && waiting[i]) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
    return 0;
}
