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;

void fail(){
    cout << "No" << endl;
    exit(0);
}

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]--;
    }
    reverse(A.begin(), A.end());
    vector<int> order, rev(M, -1);
    for(int a : A) if(rev[a] == -1){
        rev[a] = order.size();
        order.push_back(a);
    }

    int sz = order.size();
    int pt = sz-1;
    while(pt > 0 && order[pt-1] < order[pt]) pt--;
    int lack = 0;
    while(lack < M && rev[lack] >= 0) lack++;
    if(lack > order.back()) sz = pt;

    for(int a=0; a<M; a++) if(rev[a] == -1){
        rev[a] = order.size();
        order.push_back(a);
    }

    vector<int> num(M+1);
    bitset<100001> visited;
    visited[0] = 1;
    num[0] = N;
    for(int a : A){
        if(num[rev[a]]){
            num[rev[a]]--;
            num[rev[a]+1]++;
            visited[rev[a]+1] = 1;
        }else if(!visited[rev[a]]){
            fail();
        }
    }
    for(int i=0; i<sz; i++) if(num[i]){
        fail();
    }
    cout << "Yes" << endl;
    return 0;
}
