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>
#define sz(x) (int((x).size()))
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(), (x).end()
template<typename T> bool domax(T &a, T b) { return (b > a ? (a = b, true) : false); }
template<typename T> bool domin(T &a, T b) { return (b < a ? (a = b, true) : false); }
typedef long long ll;

const int maxn = 100*1000+5;

int n, m, q, a[maxn], place[maxn], next[maxn];
std::vector<int> l[maxn];
bool seen[maxn];

int main() {
    scanf("%d%d%d", &n, &m, &q);
    for (int i = 0; i < q; i++) scanf("%d", a+i);
    for (int i = q-1; i >= 0; i--) {
        int v = a[i];
        if (!seen[v]) {
            seen[v] = true;
            place[v] = sz(l[0]);
            l[next[v]++].pb(v);
        } else if (next[v] < n) {
            if (place[v] == sz(l[next[v]])) {
                l[next[v]++].pb(v);
            }
        }
    }
    std::fill(seen, seen+m+1, false);
    for (int &i : l[n-1]) seen[i] = true;
    for (int j = 1; j <= m; j++) if (!seen[j]) l[n-1].pb(j);
    for (int i = n-2; i >= 0; i--) {
        for (int k = 0; k < sz(l[i]); k++) if (l[i][k] != l[n-1][k]) {
            printf("No\n");
            exit(0);
        }
    }
    printf("Yes\n");
}
