Problem: 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 <image> replace ai with D(ai); 
  2. SUM l r — calculate <image>. 



Print the answer for each SUM query.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 3·105) — the number of elements in the array and the number of queries to process, respectively.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the elements of the array.

Then m lines follow, each containing 3 integers ti, li, ri denoting i-th query. If ti = 1, then i-th query is REPLACE li ri, otherwise it's SUM li ri (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n).

There is at least one SUM query.

Output

For each SUM query print the answer to it.

Example

Input

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


Output

30
13
4
22

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m;
int D[1000005];
int a[300005];
long long C[300005];
set<int> posset;
int lrbit(int x) { return x & -x; }
void update(int p, int x) {
  for (int i = p; i <= n; i += lrbit(i)) {
    C[i] += (x - a[p]);
  }
  a[p] = x;
}
long long Sum(int l, int r) {
  long long lsm, rsm;
  lsm = rsm = 0;
  for (int i = l - 1; i > 0; i -= lrbit(i)) {
    lsm += C[i];
  }
  for (int i = r; i > 0; i -= lrbit(i)) {
    rsm += C[i];
  }
  return rsm - lsm;
}
int main() {
  for (int i = 1; i <= 1000000; i++)
    for (int j = i; j <= 1000000; j += i) {
      ++D[j];
    }
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) {
    int t1;
    scanf("%d", &t1);
    update(i, t1);
    if (t1 != 1 && t1 != 2) posset.insert(i);
  }
  for (int i = 0; i < m; i++) {
    int o, l, r;
    scanf("%d", &o);
    scanf("%d", &l);
    scanf("%d", &r);
    if (o == 1) {
      set<int>::iterator it = posset.lower_bound(l);
      for (; it != posset.end() && (*it) <= r;) {
        update(*it, D[a[*it]]);
        if (a[*it] == 1 || a[*it] == 2) {
          posset.erase(it++);
        } else
          ++it;
      }
    } else {
      printf("%lld\n", Sum(l, r));
    }
  }
  return 0;
}