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;
const long long N = 3e5 + 5;
const long long M = 1e6 + 5;
long long ft[N], D[M], a[N];
void update(long long id, long long val) {
  while (id < N) {
    ft[id] += val;
    id += (id & -id);
  }
}
long long query(long long id) {
  long long res = 0;
  while (id > 0) {
    res += ft[id];
    id -= (id & -id);
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  ;
  for (long long i = 1; i < M; i++) {
    for (long long j = i; j < M; j += i) {
      D[j]++;
    }
  }
  long long n, q;
  cin >> n >> q;
  set<long long> active;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    update(i, a[i]);
    active.insert(i);
  }
  for (long long _i = 1; _i <= q; _i++) {
    long long type, l, r;
    cin >> type >> l >> r;
    if (type == 1) {
      set<long long> for_del;
      auto it = active.lower_bound(l);
      while (it != active.end() && *it <= r) {
        if (D[a[*it]] != a[*it]) {
          update(*it, D[a[*it]] - a[*it]);
          a[*it] = D[a[*it]];
        } else {
          for_del.insert(*it);
        }
        it++;
      }
      for (auto it : for_del) active.erase(it);
    } else {
      cout << query(r) - query(l - 1) << "\n";
    }
  }
}