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 long long INF = 1e9 + 10, MAX = 1e6 + 1e4, MOD = 1e9 + 7, MAXL = 25;
void OUT(long double o, int x) {
  cout << fixed << setprecision(x) << o;
  return;
}
long long a[MAX], c[MAX], fen[MAX], n, m;
void upd(int v, long long x) {
  for (; v <= n; v += v & -v) fen[v] += x;
}
long long g(int v) {
  long long x = 0;
  for (; v; v -= (v & -v)) x += fen[v];
  return x;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 1; i <= 1e6; i++)
    for (int j = 1; i * j <= 1e6; j++) c[i * j]++;
  cin >> n >> m;
  set<int> st;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > 2) st.insert(i);
    upd(i, a[i]);
  }
  st.insert(INF);
  while (m--) {
    long long t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      long long last = l;
      while (1) {
        long long x = *st.lower_bound(last);
        last = x + 1;
        if (x > r) break;
        long long y = g(x) - g(x - 1);
        if (c[y] < 3) st.erase(x);
        upd(x, c[y] - y);
      }
    } else
      cout << g(r) - g(l - 1) << "\n";
  }
  return 0;
}
