Problem

Given a permutation of length $ N $ $ P = \\ {P_1, P_2, \ ldots, P_N \\} $ and the integer $ K $.
Determine if the permutation $ P $ can be monotonically increased by repeating the following operation any number of times $ 0 $ or more.

* Choose the integer $ x \ (0 \ le x \ le N-K) $. Patrol right shift around $ P_ {x + 1}, \ ldots, P_ {x + K} $



However, the cyclic right shift of the subsequence $ U = U_1, \ ldots, U_M $ means that $ U = U_1, \ ldots, U_M $ is changed to $ U = U_M, U_1, \ ldots, U_ {M-1} $. Means to change.

Constraints

The input satisfies the following conditions.

* $ 2 \ leq K \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq P_i \ leq N \ (1 \ leq i \ leq N) $
* $ P_i \ neq P_j \ (i \ neq j) $
* All inputs are integers

Input

The input is given in the following format.


$ N $ $ K $
$ P_1 $ $ \ ldots $ $ P_N $


Permutation length $ N $ and integer $ K $ are given in the first row, separated by blanks.
The elements of the permutation $ P $ are given in the second row, separated by blanks.

Output

Print "Yes" if you can monotonically increase $ P $, otherwise print "No" on the $ 1 $ line.

Examples

Input

3 3
2 3 1


Output

Yes


Input

3 2
1 2 3


Output

Yes


Input

3 3
3 2 1


Output

No
#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 998244353;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef pair<ll, ll> LP;
typedef vector<ll> vec;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-5;
const ld pi = acos(-1.0);

struct BIT {
private:
	vector<ll> node; int n;
public:
	BIT(int n_) {
		n = n_; node.resize(n, 0);
	}
	//0-indexed
	void add(int a, ll w) {
		for (int i = a; i < n; i |= i + 1)node[i] += w;
	}
	//[0,a)
	ll sum(int a) {
		ll ret = 0;
		for (int i = a - 1; i >= 0; i = (i&(i + 1)) - 1)ret += node[i];
		return ret;
	}
	//[a,b)
	ll sum(int a, int b) {
		return sum(b) - sum(a);
	}
};

void solve() {
	int n, k; cin >> n >> k;
	vector<int> p(n);
	rep(i, n) {
		cin >> p[i];
	}
	if (n == k) {
		rep(i, n) {
			if (p[i] == 1) {
				rep(j, n) {
					int id = (i + j) % n;
					if (p[id] != 1 + j) {
						cout << "No" << endl; return;
					}
				}
			}
		}
		cout << "Yes" << endl;
		return;
	}
	if (k % 2 == 0) {
		cout << "Yes" << endl;
	}
	else {
		BIT b(n + 1);
		ll num = 0;
		rep(i, n) {
			num += i - b.sum(p[i]);
			b.add(p[i],1);
		}
		if (num % 2 == 0) {
			cout << "Yes" << endl;
		}
		else {
			cout << "No" << endl;

		}
	}
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	//stop
	return 0;
}


