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
#include <bits/stdc++.h>
using namespace std;
long long f[300010], d[1000010];
int n;
void add(int i, long long x) {
  while (i <= n) {
    f[i] += x;
    i += i & -i;
  }
}
long long sum(int i) {
  long long s = 0;
  while (i >= 1) {
    s += f[i];
    i -= i & -i;
  }
  return s;
}
set<int> pos;
int main() {
  int m, i, j, c, x, y;
  long long xx, val;
  for (i = 1; i <= 1000000; ++i) {
    for (j = i; j <= 1000000; j += i) {
      ++d[j];
    }
  }
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; ++i) {
    scanf("%lld", &xx);
    add(i, xx);
    if (xx > 2) pos.insert(i);
  }
  while (m--) {
    scanf("%d%d%d", &c, &x, &y);
    if (c == 1) {
      auto it = pos.lower_bound(x);
      while (it != pos.end() && (*it) <= y) {
        val = sum(*it) - sum((*it) - 1);
        add(*it, d[val] - val);
        if (d[val] <= 2)
          it = pos.erase(it);
        else
          ++it;
      }
    } else {
      printf("%lld\n", sum(y) - sum(x - 1));
    }
  }
}
