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
import math
r = lambda: int(input())
ra = lambda: [*map(int, input().split())]
p = lambda a: print('{:.6f}'.format(a))

def b(a):
    if a==1: return 1
    elif s(a): return 2
    else:
        d = {}
        c = 2
        while a!=1:
            if a%c==0:
                d[c] = 0
                while a%c==0:
                    a//=c
                    d[c]+=1
            c+=1
        an = 1
        for i in d.values(): an*=i+1
        return an
    
def s(a):
    for i in range(2, int(math.sqrt(a))+1):
        if a%i==0:
            return False
    return True

def re(a, B, c):
    for i in range(B-1, c):
        a[i] = b(a[i])
    return a
    
def su(a, b, c):
    return sum(a[b-1:c])

n, m = ra()
a = ra()
c = [ra() for i in range(m)]
for i in c:
    if i[0]==1: a = re(a,i[1],i[2])
    else: print(su(a,i[1],i[2]))
