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<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

int N, M, Q;
int A[100011];
int cnt[100011];
int rnk[100011];
int len[100011];
VI ord;

void MAIN() {
    scanf("%d%d%d", &N, &M, &Q);
    REP (i, Q) scanf("%d", A+i), A[i]--;;

    for (int i=Q; i--;) {
	int v = A[i];
	if (cnt[v] == 0) {
	    rnk[v] = ord.size();
	    ord.push_back(v);
	    len[0]++;
	    cnt[v]++; 
	} else if (cnt[v] < N && len[cnt[v]] == rnk[v]) {
	    len[cnt[v]]++;
	    cnt[v]++; 
	}
    }

    REP (i, M) if (cnt[i] == 0) ord.push_back(i);
    bool yes = true;
    for (int i=len[N-1]; i<M-1; i++) if (ord[i] > ord[i+1]) yes = false;
    puts(yes? "Yes": "No");
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}

