human
Write me a Python code to print a single integer, the answer to the query given the following:
Alice has recently received an array a_1, a_2, ..., a_n for her birthday! She is very proud of her array, and when she showed her friend Bob the array, he was very happy with her present too!

However, soon Bob became curious, and as any sane friend would do, asked Alice to perform q operations of two types on her array:

  * 1 x y: update the element a_x to y (set a_x = y). 
  * 2 l r: calculate how many non-decreasing subarrays exist within the subarray [a_l, a_{l+1}, ..., a_r]. More formally, count the number of pairs of integers (p,q) such that l ≤ p ≤ q ≤ r and a_p ≤ a_{p+1} ≤ ... ≤ a_{q-1} ≤ a_q. 


Help Alice answer Bob's queries!

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the size of the array, and the number of queries, respectively.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the elements of Alice's array.

The next q lines consist of three integers each. The first integer of the i-th line is t_i, the operation being performed on the i-th step (t_i = 1 or t_i = 2).

If t_i = 1, the next two integers are x_i and y_i (1 ≤ x_i ≤ n.
1 ≤ y_i ≤ 10^9), updating the element at position x_i to y_i (setting a_{x_i} = y_i).

If t_i = 2, the next two integers are l_i and r_i (1 ≤ l_i ≤ r_i ≤ n), the two indices Bob asks Alice about for the i-th query.

It's guaranteed that there is at least one operation of the second type.

Output

For each query of type 2, print a single integer, the answer to the query.

Example

Input

5 6
3 1 4 1 5
2 2 5
2 1 3
1 4 4
2 2 5
1 2 6
2 2 5

Output

6
4
10
7

Note

For the first query, l = 2 and r = 5, and the non-decreasing subarrays [p,q] are [2,2], [3,3], [4,4], [5,5], [2,3] and [4,5].
Solve the task in PYTHON3.
assistant
#Here is Python code to print a single integer, the answer to the query.

def naiveSolve():
    
    
    
    return


def solve():
    
    
    
    return

class SegmentTree():
    def __init__(self,arr,combinerFunction,isCommutative=True): # isCommutative means f(a,b)==f(b,a)
        self.n=len(arr)
        def cmb(a,b):
            if a==None: return b
            if b==None: return a
            return combinerFunction(a,b)
        self.cmb=cmb
        self.sz=self.n if isCommutative else pow(2,((self.n-1).bit_length())) # commutative needs power of 2 size
        self.t=[None]*(2*self.sz) # use None as initial value
        for i in range(self.n): self.t[i+self.sz]=arr[i]
        for i in range(self.sz-1,0,-1): self.pull(i)
    def pull(self,p):
        self.t[p]=self.cmb(self.t[2*p],self.t[2*p+1])
    def update(self,idx,val): # set val at idx
        idx+=self.sz
        self.t[idx]=val
        idx//=2
        while idx>0:
            self.pull(idx)
            idx//=2
    def query(self,l,r): # aggregated value in [l,r] inclusive
        l+=self.sz; r+=self.sz+1
        a=b=None
        while l<r:
            if l%2==1: a=self.cmb(a,self.t[l]); l+=1
            if r%2==1: r-=1; b=self.cmb(self.t[r],b)
            l//=2; r//=2
        return self.cmb(a,b)
        

def add(length):
    return (length*(length+1))//2
# (leftVal,rightVal,leftLen,rightLen,cntsWithin,False)
def combiner(a,b):
    alv,arv,all,arl,ac,ab=a
    blv,brv,bll,brl,bc,bb=b
    clv,crv,cll,crl,cc,cb=alv,brv,all,brl,ac+bc,ab or bb
    if ab and bb:
        if arv<=blv: cc+=add(arl+bll)
        else: cc+=add(arl)+add(bll)
    elif ab and not bb:
        if arv<=blv: crl+=arl
        else: cc+=add(arl)
    elif not ab and bb:
        if arv<=blv: cll+=bll
        else: cc+=add(bll)
    else:
        if arv<=blv: cll=crl=all+bll
        else: cb=True
    return (clv,crv,cll,crl,cc,cb)

def main():
    
    n,q=readIntArr()
    arr2=readIntArr()
    # (leftVal,rightVal,leftLen,rightLen,cntsWithin,broken)
    arr=[(x,x,1,1,0,False) for x in arr2]
    st=SegmentTree(arr,combiner,False)
    
    allans=[]
   ...