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 REP(i,a,b) for (int i = (a); i <= (b); ++i)
#define REPD(i,a,b) for (int i = (a); i >= (b); --i)
#define FORI(i,n) REP(i,1,n)
#define FOR(i,n) REP(i,0,int(n)-1)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define vi vector<int>
#define ll long long
#define SZ(x) int((x).size())
#define DBG(v) cerr << #v << " = " << (v) << endl;
#define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++)
#define fi first
#define se second

const int N = 100100;

int n,m,q,v;
int a[N], used[N], ord[N], pos[N], sor[N], prog[N];

int main() {
	scanf("%d%d%d", &n, &m, &q);
	FOR(i,q) scanf("%d", &a[i]);
	FOR(i,q) a[i]--;
	REPD(i,q-1,0) if (!used[a[i]]) {
		ord[v++] = a[i];
		used[a[i]]=1;
	}
	FOR(i,m) if (!used[i]) ord[v++] = i;
	sor[v-1]=sor[v]=1;
	REPD(i,v-2,0) sor[i] = sor[i+1] && (ord[i] < ord[i+1]);
	FOR(i,v) pos[ord[i]] = i;
	REPD(i,q-1,0) {
		int pp = upper_bound(prog, prog+n, pos[a[i]]) - prog - 1;
		if (pp >= 0 && prog[pp] == pos[a[i]]) {
			prog[pp]++;
		}
	}
	//FOR(i,n) printf("%d %d\n", i, prog[i]);
	bool ok = true;
	FOR(i,n) if (!sor[prog[i]]) ok = false;
	printf("%s\n", ok ? "Yes" : "No");
	return 0;
}
