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 REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = (int)(n) - 1; (i) >= 0; -- (i))
using namespace std;
template <class T, class U> inline void chmin(T & a, U const & b) { a = min<T>(a, b); }

bool solve(int n, int m, int q, const vector<int> & a) {
    // make the final sequence
    vector<int> used(m);
    vector<int> b;
    REP_R (i, q) {
        if (not used[a[i]]) {
            used[a[i]] = true;
            b.push_back(a[i]);
        }
    }
    REP (i, m) {
        if (not used[i]) {
            b.push_back(i);
        }
    }

    // remove the tail
    int r = m - 1;
    while (r - 1 >= 0 and b[r - 1] < b[r]) {
        -- r;
    }
    if (r == 0) return true;

    // count queries
    vector<int> next(m, m);
    REP (i, r) {
        next[b[i]] = (i + 1 < r ? b[i + 1] : m);
    }
    vector<int> cnt(m + 1);
    cnt[b[0]] = n;
    REP_R (i, q) {
        if (cnt[a[i]]) {
            -- cnt[a[i]];
            ++ cnt[next[a[i]]];
        }
    }
    return cnt[m] == n;
}

int main() {
    int n, m; cin >> n >> m;
    int q; cin >> q;
    vector<int> a(q);
    REP (i, q) {
        cin >> a[i];
        -- a[i];
    }
    cout << (solve(n, m, q, a) ? "Yes" : "No") << endl;
    return 0;
}
