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 MAXN = 300000;
int n, q;
int a;
int t, l, r;
map<int, queue<int>> arr;
class BIT {
 private:
  long long arr[MAXN + 1];
  inline int lb(int a) { return a & -a; }

 public:
  void mdy(int n, int a) {
    for (int i = n; i <= MAXN; i += lb(i)) arr[i] += a;
  }
  long long qry(int n) {
    long long sum = 0;
    for (int i = n; i; i -= lb(i)) sum += arr[i];
    return sum;
  }
} bit;
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    arr[i].push(a);
    bit.mdy(i, a);
    while (a > 2) {
      int c = 0;
      for (int j = 1; j * j <= a; j++)
        if (a % j == 0) c += 1 + (a / j != j);
      a = c;
      arr[i].push(a);
    }
  }
  for (int i = 0; i < q; i++) {
    cin >> t >> l >> r;
    if (t == 1) {
      auto it = arr.lower_bound(l);
      while (it != arr.end() && it->first <= r) {
        int tmp = it->second.front();
        it->second.pop();
        if (!it->second.size()) {
          it++;
          arr.erase(prev(it));
          continue;
        }
        bit.mdy(it->first, it->second.front() - tmp);
        it++;
      }
    } else {
      cout << bit.qry(r) - bit.qry(l - 1) << "\n";
    }
  }
}
