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 <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;

set<int> st[100000];
bool used[100000];
int idx[100000];

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    int N, M, Q;
    cin >> N >> M >> Q;
    int a[100000];
    for(int i = 0; i < Q; i++){
        cin >> a[i];
        a[i]--;
        st[a[i]].insert(i);
    }
    vector<int> ans;
    for(int i = Q-1; i >= 0; i--){
        if(!used[a[i]]){
            ans.push_back(a[i]);
            used[a[i]] = true;
        }
    }
    for(int i = 0; i < N; i++) idx[i] = Q;
    for(int i = 0; i < M; i++){
        if(!used[i]) ans.push_back(i);
    }
    //for(int i = 0; i < M; i++) cout << ans[i] << ' ';
    int last = -1;
    for(int i = 1; i < M; i++){
        if(ans[i] < ans[i-1]) last = i-1;
    }
    // cout << last << endl;
    for(int i = 0; i <= last; i++){
        //cout << i << endl;
        for(int j = 0; j < N; j++){
            //cout << j << endl;
            auto ptr = st[ans[i]].lower_bound(idx[j]);
            if(ptr == st[ans[i]].begin()){
                cout << "No" << endl;
                return 0;
            }
            ptr--;
            int m = *ptr;
            idx[j] = m;
            st[ans[i]].erase(m);
        }
    }
    cout << "Yes" << endl;
}