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 <vector>
#include <stack>
#include <queue>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <set>

using namespace std;

typedef long long ll;
typedef pair<int,int> PP;

/*
    freopen("input","r",stdin);
    freopen("output","w",stdout);
*/

int main() {
    ios::sync_with_stdio(false);
    int N, M, Q;
    cin >> N >> M >> Q;
    unordered_map<int, int> X;
    vector<int> order;
    vector<int> num;
    stack<int> S;
    unordered_set<int> us;
    int a, b, c;
    for (int q = 0;q < Q;q++) {
    	cin >> a;
    	S.push(a);
    }
    while (S.size() > 0) {
    	a = S.top();
    	S.pop();
    	if (X.find(a) == X.end()) {
    		b = X.size();
    		X[a] = b;
    	}
    	b = X[a];
    	if (b >= order.size()) {
    		order.push_back(a);
    		us.insert(a);
    	}
    	if (b >= num.size()) num.push_back(0);
    	if (b == 0 && num[0] < N) num[0]++;
    	else {
    		if (num[b - 1] > num[b]) num[b]++;
    	}
    }
    vector<int> rem;
    for (int i = 0;i < order.size();i++) {
    	if (num[i] != N) rem.push_back(order[i]);
    }
    for (int i = 1;i <= M;i++) {
    	if (us.find(i) == us.end()) rem.push_back(i);
    }
    vector<int> g = rem;
    sort(g.begin(), g.end());
    if (g == rem) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}