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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
    int n, m; cin>>n>>m;
    int q; cin>>q;
    int a[100010];
    for(int i=0; i<q; i++){
        cin>>a[i];
    }
    int ind[100010];
    fill(ind, ind+m+1, -1);
    int cnt[100010]={};
    vector<int> v;
    for(int i=q-1; i>=0; i--){
        if(ind[a[i]]==-1){
            ind[a[i]]=cnt[0];
            v.push_back(a[i]);
            cnt[0]++;
        }else{
            int l=0, r=n;
            while(r-l>1){
                int mid=(l+r)/2;
                if(cnt[mid]<=ind[a[i]]) r=mid;
                else l=mid;
            }
            if(r<n && cnt[r]==ind[a[i]]) cnt[r]++;
        }
    }
    for(int i=1; i<=m; i++) if(ind[i]==-1) v.push_back(i);
    int mn=0;
    for(int i=m-1; i>=1; i--){
        if(v[i]<v[i-1]){
            mn=i; break;
        }
    }
    for(int i=0; i<n; i++){
        if(cnt[i]<mn){
            cout<<"No"<<endl; return 0;
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}