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;
const int MAX_N = 524288;
const int MAX_A = 1000000;
int d[MAX_A + 1];
long long Sum[MAX_N * 2];
set<int> Set;
void update(int x, long long key) {
  x += MAX_N;
  Sum[x] = key;
  for (x /= 2; x; x /= 2) {
    Sum[x] = Sum[x + x] + Sum[x + x + 1];
  }
}
long long ask(int l, int r) {
  long long ret = 0;
  l += MAX_N;
  r += MAX_N;
  for (; l <= r; l /= 2, r /= 2) {
    if (l % 2 == 1) ret += Sum[l++];
    if (r % 2 == 0) ret += Sum[r--];
  }
  return ret;
}
int main() {
  for (int i = 1; i <= MAX_A; i++)
    for (int j = i; j <= MAX_A; j += i) {
      d[j]++;
    }
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%I64d", &x);
    update(i, x);
    if (x > 2) Set.insert(i);
  }
  for (int i = 0; i < m; i++) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      vector<int> to_erase;
      to_erase.clear();
      for (auto it = Set.lower_bound(l); it != Set.upper_bound(r); ++it) {
        update(*it, d[Sum[*it + MAX_N]]);
        if (Sum[*it + MAX_N] <= 2) to_erase.push_back(*it);
      }
      for (int j = 0; j < to_erase.size(); j++) {
        Set.erase(to_erase[j]);
      }
    } else {
      cout << ask(l, r) << endl;
    }
  }
}
