Snuke has a permutation (P_0,P_1,\cdots,P_{N-1}) of (0,1,\cdots,N-1).

Now, he will perform the following operation exactly once:

* Choose K consecutive elements in P and sort them in ascending order.



Find the number of permutations that can be produced as P after the operation.

Constraints

* 2 \leq N \leq 200000
* 2 \leq K \leq N
* 0 \leq P_i \leq N-1
* P_0,P_1,\cdots,P_{N-1} are all different.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
P_0 P_1 \cdots P_{N-1}


Output

Print the number of permutations that can be produced as P after the operation.

Examples

Input

5 3
0 2 1 4 3


Output

2


Input

4 4
0 1 2 3


Output

1


Input

10 4
2 0 1 3 7 5 4 6 8 9


Output

6
n,k = map(int,input().split())
p = tuple(map(int,input().split()))

class Seg_min():
    def __init__(self,x):
        #####単位元######
        self.ide_ele_min = 10**10
        self.func = min

        self.n = len(x)

        #num_max:n以上の最小の2のべき乗
        self.num_max =2**(self.n-1).bit_length()
        self.x = [self.ide_ele_min]*2*self.num_max

        for i,num in enumerate(x, self.num_max):
            self.x[i] = num
        for i in range(self.num_max-1,0,-1):
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def update(self,i,x):
        i += self.num_max
        self.x[i] = x
        while(x>0):
            i = i//2
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def query(self,i,j):
        res = self.ide_ele_min
        if i>=j:
            return res
        i += self.num_max
        j += self.num_max -1
        while(i<=j):
            if(i==j):
                res = self.func(res,self.x[i])
                break
            if(i&1):
                res = self.func(res,self.x[i])
                i += 1
            if(not j&1):
                res = self.func(res,self.x[j])
                j -= 1
            i = i>>1
            j = j>>1
        return res

class Seg_max():
    def __init__(self,x):
        #####単位元######
        self.ide_ele_min = 10**10 * -1
        self.func = max

        self.n = len(x)

        #num_max:n以上の最小の2のべき乗
        self.num_max =2**(self.n-1).bit_length()
        self.x = [self.ide_ele_min]*2*self.num_max

        for i,num in enumerate(x, self.num_max):
            self.x[i] = num
        for i in range(self.num_max-1,0,-1):
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def update(self,i,x):
        i += self.num_max
        self.x[i] = x
        while(x>0):
            i = i//2
            self.x[i] = self.func(self.x[i<<1],self.x[(i<<1) + 1])

    def query(self,i,j):
        res = self.ide_ele_min
        if i>=j:
            return res
        i += self.num_max
        j += self.num_max -1
        while(i<=j):
            if(i==j):
                res = self.func(res,self.x[i])
                break
            if(i&1):
                res = self.func(res,self.x[i])
                i += 1
            if(not j&1):
                res = self.func(res,self.x[j])
                j -= 1
            i = i>>1
            j = j>>1
        return res

seg_min = Seg_min(p)
seg_max = Seg_max(p)

natural = 0
ans = 0

head = 0
for i in range(1,n):
    if(p[i]<p[i-1]):
        head = i

    if(i-head >= k-1):
        natural = 1
        continue

    if(i==k-1):
        ans +=1
    if(i<k):
        continue
    ans += 1

    left = seg_min.query(i-k,i)
    right = seg_max.query(i-k+1,i+1)
    if(left==p[i-k])&(right==p[i]):
        ans -= 1

print(ans+natural)