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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
#include <complex>
#include <array>
using namespace std;
 
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
typedef vector<double> VD;
typedef vector<VD> VVD;

template<typename T> void chmin(T &a, T b) { if (a > b) a = b; }
template<typename T> void chmax(T &a, T b) { if (a < b) a = b; }

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

int main() {
    int n, m, q;
    cin >> n >> m >> q;
    VI a(q);
    REP(i,q) a[i] = in() - 1;
    VI p;
    set<int> st;
    REP(i,m) st.insert(i);
    FORR(i,q-1,0){
        if (st.count(a[i])){
            p.push_back(a[i]);
            st.erase(a[i]);
        }
    }
    for (int x : st) p.push_back(x);
    VI idx(m);
    REP(i,m) idx[p[i]] = i;

    // REP(i,m) cout << p[i] << " ";
    // cout << endl;

    int x = m - 1;
    while (x > 0 && p[x - 1] < p[x]) x--;

    VI cnt(m);
    FORR(i,q-1,0){
        int j = idx[a[i]];
        cnt[j] = min(n, cnt[j] + 1);
        if (j) chmin(cnt[j], cnt[j - 1]);
    }
    REP(i,x){
        if (cnt[i] != n){
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
    return 0;
}