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;

#define pb push_back
#define mp make_pair
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
typedef long long LL;
typedef pair<int, int> PII;

int n, m, q;
int a[100005];
int b[200005];
int last[200005];
bool used[200005];
int cnt;

void corr(int *a, int &cnt) {
    memset(last, -1, sizeof last);
    forn(i, cnt) used[i] = true;
    forn(i, cnt) {
        if (last[a[i]] != -1) {
            used[last[a[i]]] = false;
        }
        last[a[i]] = i;
    }
    int ncnt = 0;
    forn(i, cnt) if (used[i]) {
        a[ncnt++] = a[i];
    }
    cnt = ncnt;
    reverse(a, a + cnt);
    forn(i, n) if (last[i] == -1) {
        a[cnt++] = i;
    }
}

int pos[100000] = {};
set<int> where[100000];

int main() {
    //freopen("input.txt", "r", stdin);
    scanf("%d%d%d", &n, &m, &q);
    forn(i, q) scanf("%d", a + i), --a[i];
    forn(i, m) b[i] = m - 1 - i;
    cnt = m;
    forn(i, q) b[cnt++] = a[i];
    corr(b, cnt);
    while (cnt > 1 && b[cnt - 1] > b[cnt - 2]) {
        --cnt;
    }
    --cnt;
    if (cnt == 0) {
        cout << "Yes" << endl;
        return 0;
    }
    reverse(a, a + q);
    forn(i, n) {
        pos[i] = 0;
        where[b[0]].insert(i);
    }
    forn(i, q) {
        int val = a[i];
        if (!where[val].empty()) {
            int ind = *where[val].begin();
            where[val].erase(where[val].begin());
            ++pos[ind];
            if (pos[ind] < cnt) {
                where[b[pos[ind]]].insert(ind);
            }
        }
    }
    forn(i, n) if (pos[i] != cnt) {
        cout << "No" << endl;
        return 0;
    }
    cout << "Yes" << endl;
    return 0;
}
