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 <stdio.h>
# include <bits/stdc++.h>
using namespace std;
# define fi cin
# define fo cout
# define x first
# define y second
# define ll long long
# define IOS ios_base :: sync_with_stdio(0);cin.tie(0)
# define p(v) cerr << #v << " = " << v << '\n'
# define p2(v) cerr << #v << " = " << (complex < int > (v.x,v.y)) << '\n'
# define vi vector < int >
# define vll vector < ll >
# define pii pair < int , int >
# define mp make_pair
int main(void)
{
    #ifdef CF
    freopen("input","r",stdin);
    #endif // CF
    srand(time(0));
    fo << fixed << setprecision(7);
    cerr << fixed << setprecision(7);
    int n,m,q;
    static int s[1 << 20];
    fi>>n>>m>>q;
    for (int i = 1;i <= q;++i)
        fi>>s[i];
    vector < int > S;
    static int was[1 << 20];
    for (int i = 1;i <= m;++i)
        was[i] = -1;
    for (int i = q;i;--i)
        if (was[s[i]] == -1)
            S.push_back(s[i]),was[s[i]] = 1;
    static int cnt[1 << 20];
    for (int i = 1;i <= m;++i)
        if (was[i] == -1)
            S.push_back(i);
    static int where[1 << 20];
    for (int i = 0;i < m;++i)
        where[S[i]] = i + 1;
    cnt[0] = n;
    for (int i = q;i;--i)
    {
        const int index = where[s[i]] - 1;
        if (cnt[index])
            --cnt[index],++cnt[index + 1];
    }
    int Last = 0;
    for (int i = m - 1;i;--i)
        if (S[i] < S[i - 1])
            {Last = i;break;}
    int ok = 1;
    for (int i = 0;i < Last;++i)
        ok &= !cnt[i];
    fo << (ok ? "Yes" : "No") << '\n';
    cerr << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms" << '\n';
    return 0;
}