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 long long MOD = 1000000007;
const int INF = 1<<29;
const double EPS = 1e-8;
typedef vector<int> vec;
typedef pair<int,int> P;
typedef long long ll;
 
int main(){
    int N, M, Q;
    cin >> N >> M >> Q;
 
    vec a(Q);
    for(int i = 0; i < Q; i++){
        cin >> a[i];
        a[i]--;
    }
 
    vector<bool> used(M, false);
    vec que(M + 1, 0);
    vec next(M, -1);
    int prev = -1;
 
    que[M] = N - 1;
 
    for(int p = Q - 1; p >= 0; p--){
        if (que[a[p]] > 0){
            que[a[p]]--;
            if(next[a[p]] != -1){
                assert(0 <= next[a[p]] && next[a[p]] < M);
                que[next[a[p]]]++;
            }else{
                que[M]++;
            }
        }else{
            if(!used[a[p]]){
                used[a[p]] = true;
                if(prev >= 0) next[prev] = a[p];
                prev = a[p];
                que[a[p]] += que[M];
                que[M] = 0;
            }
        }
    }
 
    for(int i = 0; i < M; i++){
        if(!used[i]){
            used[i] = true;
            if(prev >= 0) next[prev] = i;
            prev = i;
            que[i] += que[M];
            que[M] = 0;
        }
        if(que[i] > 0){
            if(next[i] != -1){
                assert(0 <= next[i] && next[i] < M);
                que[next[i]] += que[i];
            }else{
                que[M] += que[i];
            }
        }
    }
 
    if(que[M] == N - 1){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
    return 0;
}