Let *D*(*x*) be the number of positive divisors of a positive integer *x*. For example, *D*(2) = 2 (2 is divisible by 1 and 2), *D*(6) = 4 (6 is divisible by 1, 2, 3 and 6).

You are given an array *a* of *n* integers. You have to process two types of queries:
 1.  REPLACE *l* *r* — for every <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6afb5e3d414ffbb040c175e23c6627da949add06.png" style="max-width: 100.0%;max-height: 100.0%;"/> replace *a*<sub class="lower-index">*i*</sub> with *D*(*a*<sub class="lower-index">*i*</sub>); 1.  SUM *l* *r* — calculate <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1e50b9417bb7e50d92d43134922596e938249431.png" style="max-width: 100.0%;max-height: 100.0%;"/>. 
Print the answer for each SUM query.
The first line contains two integers *n* and *m* (1 ≤ *n*, *m* ≤ 3·10<sup class="upper-index">5</sup>) — the number of elements in the array and the number of queries to process, respectively.

The second line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">6</sup>) — the elements of the array.

Then *m* lines follow, each containing 3 integers *t*<sub class="lower-index">*i*</sub>, *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub> denoting *i*-th query. If *t*<sub class="lower-index">*i*</sub> = 1, then *i*-th query is REPLACE *l*<sub class="lower-index">*i*</sub> *r*<sub class="lower-index">*i*</sub>, otherwise it's SUM *l*<sub class="lower-index">*i*</sub> *r*<sub class="lower-index">*i*</sub> (1 ≤ *t*<sub class="lower-index">*i*</sub> ≤ 2, 1 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ *n*).

There is at least one SUM query.
none
7 6
6 4 1 10 3 2 4
2 1 7
2 4 5
1 3 5
2 4 4
1 5 7
2 1 7
30
13
4
22
4 2
1 1 1 3
1 1 4
2 1 4
5
10 2
1 1 1 1 1 1 1 1 1 9
1 1 10
2 1 10
12
4 2
1 1 3 1
1 1 4
2 1 4
5
5 3
1 1 1 1 3
2 1 5
1 1 5
2 1 5
7
6
7 10
1 1 1 1 1 1 1
1 1 7
1 1 7
1 1 7
1 1 7
1 1 7
1 1 7
1 1 7
1 1 7
1 1 7
2 1 7
7
def fun(n):
    sqrt=int(n**0.5)
    ct=0
    for t in range(1,sqrt+1):
        if(n % t==0):
            if(n/t==t):
                ct+=1
            else:
                ct+=2
    return ct


n,m=[int(i) for i in input().split()]

a=[int(i) for i in input().split()]

b=[[int(i) for i in input().split()] for i in range(m)]


for i in range(m):
    s=0
    if(b[i][0]==2):
        for j in range(b[i][1]-1,b[i][2]):
            s+=a[j]
        print(s)
    elif(b[i][0]==1):
        for j in range(b[i][1]-1,b[i][2]):
            a[j]=fun(a[j])


