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 M = 300007;
const int MM = 1e6 + 7;
const long double esl = 1e-5;
long long sa[MM], a[M];
pair<long long, int> ST[5 * M];
void sang() {
  for (int i = 1; i < MM; i++)
    for (int j = i; j < MM; j += i) sa[j]++;
}
void build(int l, int r, int id) {
  if (l == r) {
    ST[id] = make_pair(a[l], a[l]);
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, id * 2);
  build(mid + 1, r, id * 2 + 1);
  ST[id].first = ST[id * 2].first + ST[id * 2 + 1].first;
  ST[id].second = max(ST[id * 2].second, ST[id * 2 + 1].second);
}
void update(int l, int r, int u, int v, int id) {
  if (u > r || v < l) return;
  if (ST[id].second <= 2) return;
  if (u == v) {
    ST[id].first = ST[id].second = sa[ST[id].first];
    return;
  }
  int mid = (u + v) / 2;
  update(l, r, u, mid, id * 2);
  update(l, r, mid + 1, v, id * 2 + 1);
  ST[id].first = ST[id * 2].first + ST[id * 2 + 1].first;
  ST[id].second = max(ST[id * 2].second, ST[id * 2 + 1].second);
}
long long get(int l, int r, int u, int v, int id) {
  if (u > r || v < l) return 0;
  if (u >= l && v <= r) return ST[id].first;
  int mid = (u + v) / 2;
  return get(l, r, u, mid, id * 2) + get(l, r, mid + 1, v, id * 2 + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int t, l, r;
  int n, m;
  sang();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  while (m--) {
    cin >> t >> l >> r;
    if (t == 1)
      update(l, r, 1, n, 1);
    else
      cout << get(l, r, 1, n, 1) << endl;
  }
  return 0;
}