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>

using namespace std;

typedef long long llint;
typedef pair <int, int> pi;

const int MAXN = 100005;

int n, m, q;
int a[MAXN], bio[MAXN], cnt[MAXN], nxt[MAXN];
vector <int> v, r;

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m >> q;
    for (int i = 0; i < q; i++) {
        cin >> a[i];
    }
    for (int i = q-1; i >= 0; i--) {
        if (!bio[a[i]]) v.push_back(a[i]);
        bio[a[i]] = 1;
    }
    for (int i = 1; i <= m; i++) if (bio[i] == 0) v.push_back(i);
    for (int i = 1; i < m; i++) nxt[v[i - 1]] = v[i];
    cnt[v[0]] = n;
    for (int i = q-1; i >= 0; i--) {
        if (cnt[a[i]] == 0) continue;
        cnt[a[i]]--;
        cnt[nxt[a[i]]]++;
    }
    memset(bio, 0, sizeof bio);
    for (int i = 0; i < m; i++) {
        if (cnt[v[i]] != 0) break;
        bio[v[i]] = 1;
        r.push_back(v[i]);
    }
    for (int i = 1; i <= m; i++) if (bio[i] == 0) r.push_back(i);
    bool ok = 1;
    for (int i = 0; i < m; i++) ok &= v[i] == r[i];
    if (ok) cout << "Yes"; else cout << "No";
    return 0;
}
