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;
long long n, m, a[300005], d[1000006];
struct seg_tree {
  long long ma[2000006], su[2000006];
  void build(long long id, long long l, long long r) {
    if (l == r) {
      ma[id] = su[id] = a[l];
      return;
    }
    long long mid = (l + r) >> 1;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    ma[id] = max(ma[id * 2], ma[id * 2 + 1]);
    su[id] = su[id * 2] + su[id * 2 + 1];
  }
  void upd(long long id, long long l, long long r, long long u, long long v) {
    if (l > v || r < u) return;
    if (u <= l && r <= v && ma[id] <= 2) return;
    if (l == r) {
      ma[id] = d[ma[id]];
      su[id] = d[su[id]];
      return;
    }
    long long mid = (l + r) >> 1;
    upd(id * 2, l, mid, u, v);
    upd(id * 2 + 1, mid + 1, r, u, v);
    su[id] = su[id * 2] + su[id * 2 + 1];
    ma[id] = max(ma[id * 2], ma[id * 2 + 1]);
  }
  long long get(long long id, long long l, long long r, long long u,
                long long v) {
    if (l > v || r < u) return 0;
    if (u <= l && r <= v) return su[id];
    long long mid = (l + r) >> 1;
    return get(id * 2, l, mid, u, v) + get(id * 2 + 1, mid + 1, r, u, v);
  }
} seg;
void Solve() {
  cin >> n >> m;
  long long tmp = 0;
  for (long long i = 1, _b = n; i <= _b; i++) cin >> a[i], tmp = max(tmp, a[i]);
  for (long long i = 1, _b = tmp; i <= _b; i++) {
    for (long long j = i; j <= tmp; j += i) d[j]++;
  }
  seg.build(1, 1, n);
  while (m--) {
    long long t, l, r;
    cin >> t >> l >> r;
    if (t == 1)
      seg.upd(1, 1, n, l, r);
    else
      cout << seg.get(1, 1, n, l, r) << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int Test_numbers = 1;
  while (Test_numbers--) Solve();
  return 0;
}