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 int N = 3e+5 + 5;
const long long inf = 4e+18;
const long long mod = 1e+9 + 7;
inline void boost() {
  ios_base ::sync_with_stdio(NULL);
  cin.tie(NULL), cout.tie(NULL);
}
struct vert {
  long long sum, mx;
};
vert t[1201200];
long long kol[1001000], a[300300];
void build(int v, int tl, int tr) {
  if (tl == tr)
    t[v].sum = t[v].mx = a[tl];
  else {
    int tm = (tl + tr) / 2;
    build(v * 2, tl, tm);
    build(v * 2 + 1, tm + 1, tr);
    t[v].sum = t[v * 2].sum + t[v * 2 + 1].sum;
    t[v].mx = max(t[v * 2].mx, t[v * 2 + 1].mx);
  }
}
void update(int v, int tl, int tr, int l, int r) {
  if (l > r) return;
  if (tl == tr) {
    int first = (int)t[v].sum;
    t[v].sum = t[v].mx = kol[first];
    return;
  }
  int tm = (tl + tr) / 2;
  if (t[v * 2].mx > 2) update(v * 2, tl, tm, l, min(tm, r));
  if (t[v * 2 + 1].mx > 2) update(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r);
  t[v].sum = t[v * 2].sum + t[v * 2 + 1].sum;
  t[v].mx = max(t[v * 2].mx, t[v * 2 + 1].mx);
}
long long get(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0ll;
  if (tl == l && tr == r) return t[v].sum;
  int tm = (tl + tr) / 2;
  return (get(v * 2, tl, tm, l, min(tm, r)) +
          get(v * 2 + 1, tm + 1, tr, max(tm + 1, l), r));
}
int main() {
  boost();
  for (int i = 1; i <= 1000000; i++)
    for (int j = i; j <= 1000000; j += i) kol[j]++;
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  build(1, 0, n - 1);
  for (int i = 0; i < m; i++) {
    int t, l, r;
    cin >> t >> l >> r;
    l--, r--;
    if (t == 1)
      update(1, 0, n - 1, l, r);
    else
      cout << get(1, 0, n - 1, l, r) << "\n";
  }
  return 0;
}