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;
set<pair<pair<int, int>, int> > c;
int Arr[300002];
long long seg[1200003];
int p[1000002];
int n;
void upd(int x, int d) {
  for (; x <= n; x += x & -x) seg[x] += d;
}
long long qu(int x) {
  long long ans = 0;
  for (; x > 0; x -= x & -x) ans += seg[x];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) p[j]++;
  }
  int q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> Arr[i];
    upd(i + 1, Arr[i]);
    c.insert({{i, Arr[i]}, 6});
  }
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      pair<pair<int, int>, int> s = {{l - 1, -1}, -1};
      auto x = c.lower_bound(s);
      while (x->first.first < r && x != c.end()) {
        int val = x->first.second;
        int k = x->second;
        int ind = x->first.first;
        x++;
        int cnt = p[val];
        c.erase({{ind, val}, k});
        upd(ind + 1, cnt - val);
        if (k - 1) c.insert({{ind, cnt}, k - 1});
      }
    } else
      cout << qu(r) - qu(l - 1) << endl;
  }
  return 0;
}
