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
// ※※※ 解答不能 ※※※
// square1001氏.
// https://atcoder.jp/contests/code-festival-2016-quala/submissions/1939048
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using P = pair<int, int>;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
int N, M, Q, a[100009], c[100009], p[100009], inv[100009];
int main(){
    scanf("%d%d%d", &N, &M, &Q);
    rep(i, Q) scanf("%d", &a[i]), a[i]--;
    rep(i, M) inv[i] = -1;
    int cur = 0;
    repr(i, Q - 1, 0){
        if(inv[a[i]] == -1){
            inv[a[i]] = cur;
            c[cur] = 1;
            p[cur++] = a[i];
        }else{
            if(c[inv[a[i]]] < (inv[a[i]] == 0 ? N : c[inv[a[i]] - 1])) c[inv[a[i]]]++;
        }
    }
    rep(i, M) if(inv[i] == -1) p[cur++] = i;
    bool ret = true;
    rep(i, M) if(i >= 1 && c[i - 1] != N && p[i - 1] > p[i]) ret = false;
    printf(ret ? "Yes\n" : "No\n");
    return 0;
}