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>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long mod = 1e9 + 7;
long long n, m, t, l, r;
long long b[300001];
long long a[300001];
long long d[1000001];
set<long long> s;
void Upd(long long p, long long v) {
  while (p <= n) {
    b[p] += v;
    p += (p & -p);
  }
}
long long Qry(long long nod) {
  long long ans = 0;
  while (nod) {
    ans += b[nod];
    nod -= (nod & -nod);
  }
  return ans;
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  for (long long i = 1; i <= 1e6; i++) {
    for (long long j = i; j <= 1e6; j += i) d[j]++;
  }
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    Upd(i, a[i]);
    if (a[i] > 2) s.insert(i);
  }
  while (m--) {
    cin >> t >> l >> r;
    if (t == 2)
      cout << Qry(r) - Qry(l - 1) << '\n';
    else {
      auto it = s.lower_bound(l);
      while (it != s.end() && *it <= r) {
        long long xd = *it;
        if (d[a[xd]] <= 2)
          s.erase(it++);
        else
          it++;
        Upd(xd, d[a[xd]] - a[xd]);
        a[xd] = d[a[xd]];
      }
    }
  }
}