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 <bits/stdc++.h>
using namespace std;

template<typename T>
struct BIT {
    int n;
    vector<T> dat;

    BIT(int n=0){
        initialize(n);
    }

    void initialize(int nin){
        n = nin;
        dat.resize(n, 0);
    }

    T sum(int i){
        T s = 0;
        while(i >= 0){
            s += dat[i];
            i = (i & (i+1)) - 1;
        }
        return s;
    }

    T sum_between(int i, int j){
        return sum(j) - sum(i-1);
    }

    void plus(int i, T x){
        while(i < n){
            dat[i] += x;
            i |= i+1;
        }
    }

    // a[0]+...+a[ret] >= x
    int lower_bound(T x){
        int ret = -1;
        int k = 1;
        while(2*k <= n) k <<= 1;
        for( ;k>0; k>>=1){
            if(ret+k < n && dat[ret+k] < x){
                x -= dat[ret+k];
                ret += k;
            }
        }
        return ret + 1;
    }
};

int main() {
    int N, K, P[100000];
    cin >> N >> K;
    for(int i=0; i<N; i++) cin >> P[i];

    bool ok = true;
    if(K == N){
        for(int i=0; i<N; i++) if((N+P[(i+1)%N]-P[i])%N != 1) ok = false;
    }else{
        int64_t s = 0;
        BIT<int> bit(N+1);
        for(int i=0; i<N; i++){
            s += bit.sum_between(P[i], N);
            bit.plus(P[i], 1);
        }
        ok = !(K%2) || !(s%2);
    }
    cout << (ok ? "Yes" : "No") << endl;
    return 0;
}
