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 i, j;
const int M = 3e5 + 6;
const int MAX = 1e6 + 6;
long long arr[M];
long long bit[M];
long long D[MAX];
void add(int idx, long long val, int n) {
  long long del = val - arr[idx];
  for (; idx < n; idx = idx | (idx + 1)) {
    bit[idx] += del;
  }
}
long long sum(int r) {
  long long ret = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    ret += bit[r];
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  for (i = 1; i < MAX; i++) {
    for (j = i; j < MAX; j += i) D[j]++;
  }
  int n, m;
  cin >> n >> m;
  set<int> ids;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    add(i, 2 * arr[i], n);
    if (arr[i] > 2) ids.insert(i);
  }
  while (m--) {
    int t, l, r;
    cin >> t >> l >> r;
    l--, r--;
    if (t == 2) {
      cout << sum(r) - sum(l - 1) << "\n";
    } else {
      auto lo = ids.lower_bound(l);
      auto hi = ids.upper_bound(r);
      for (auto it = lo; it != hi;) {
        int id = *it;
        long long x = D[arr[id]];
        add(id, x, n);
        arr[id] = x;
        if (x <= 2)
          it = ids.erase(it);
        else
          it++;
      }
    }
  }
}