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 <cstdio>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define REP(i,n) for (int i=1;i<=(n);++i)
#define FOR(i,a,b) for (int i=(a);i<=(b);++i)
#define ROF(i,a,b) for (int i=(a);i>=(b);--i)
#define FEC(p,u) for (edge*p=head[u];p;p=p->nxt)
using namespace std;
typedef long long LL;

int n, m, q, a[110000];
int cur[110000], pre[110000], pr[110000], nx[110000];
bool vst[110000], ans = true;
int main(){
	scanf("%d%d%d", &n, &m, &q);
	REP(i,m) cur[i] = -i;
	REP(i,q) {
		scanf("%d", a+i);
		pre[i] = cur[a[i]];
		cur[a[i]] = i;
	}
	REP(i,m) cur[i] = q+1;
	REP(i,m) pr[i] = i-1, nx[i] = i+1;
	int step(0);
	ROF(i,q,1) if (!vst[a[i]]) {
		vst[a[i]] = true;
		nx[pr[a[i]]] = nx[a[i]], pr[nx[a[i]]] = pr[a[i]];
		++step;
		int k = 1, j = i;
		while (k <= n) {
			while (j > 0 && j > cur[k]) j = pre[j];
			if (cur[k] < 0 && j != -nx[-cur[k]]) ans = false;
			if (cur[k] < 0) { cur[k] = j; break; }
			else cur[k] = j;
			++k; if (j > 0) j = pre[j];
		}
	}
	REP(i,m) if (cur[i]<-step) ans = false;
	if (ans) printf("Yes\n"); else printf("No\n");
	return 0;
}
