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;

using i64 = int64_t;

const i64 MOD = 1e9+7;

const i64 INF = 1e18+7;


signed main(){
    int n, m;
    cin >> m >> n;
    int q;
    cin >> q;
    vector<int> a(q);
    for(auto& x : a){
        cin >> x;
        --x;
    }

    auto rev = a;
    reverse(rev.begin(), rev.end());
    for(int i = 0; i < n; ++i)
        rev.emplace_back(i);
    unordered_set<int> s;
    vector<int> final;
    for(auto& x : rev)
        if(s.find(x) == s.end()){
            s.insert(x);
            final.emplace_back(x);
        }
    rev.resize(rev.size() - n);

    vector<int> inv(n);
    for(int i = 0; i < n; ++i)
        inv[final[i]] = i;

    vector<int> dp(n + 1, 0);
    dp[0] = m;
    int t = 0;
    for(auto& x : rev){
        if(t < inv[x]){
            cout << "No" << endl;
            return 0;
        }
        if(dp[inv[x]]){
            --dp[inv[x]];
            ++dp[inv[x] + 1];
        }
        t = max(t, inv[x] + 1);
    }

    int c;
    int ma = MOD;
    for(c = n - 1; c >= 0; --c){
        if(ma < final[c])
            break;
        ma = final[c];
    }
    c++;

    /*
    cout << c << endl;

    for(int i = 0; i < n; ++i)
        cout << final[i] + 1 << " \n"[i == n - 1];

    for(int i = 0; i <= n; ++i)
        cout << dp[i] << " \n"[i == n];
        */

    cout << (accumulate(next(dp.begin(), c), dp.end(), 0) == m ? "Yes" : "No") << endl;
}
