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 int N = 1e6 + 5;
int p[N], mx[4 * N], a[N];
long long t[4 * N];
void update(int l, int x, int tl, int tr, int v = 1) {
  if (tl == tr) {
    t[v] = x;
    mx[v] = x;
    return;
  }
  int tm = (tl + tr) >> 1;
  if (tm >= l)
    update(l, x, tl, tm, v + v);
  else
    update(l, x, tm + 1, tr, v + v + 1);
  t[v] = t[v + v] + t[v + v + 1];
  mx[v] = max(mx[v + v], mx[v + v + 1]);
}
vector<int> arr;
void fnd(int l, int r, int tl, int tr, int v = 1) {
  if (tr < l || tl > r || mx[v] <= 2) {
    return;
  }
  if (tl == tr) {
    arr.push_back(tl);
    return;
  }
  int tm = (tl + tr) >> 1;
  fnd(l, r, tl, tm, v + v);
  fnd(l, r, tm + 1, tr, v + v + 1);
}
long long get(int l, int r, int tl, int tr, int v = 1) {
  if (tl > r || tr < l) {
    return 0;
  }
  if (l <= tl && tr <= r) {
    return t[v];
  }
  int tm = (tl + tr) >> 1;
  return get(l, r, tl, tm, v + v) + get(l, r, tm + 1, tr, v + v + 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int c = 0;
  for (int i = 2; i <= 1000000; i++) {
    p[i]++;
    for (int j = i; j <= 1000000; j += i) {
      p[j]++;
    }
  }
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    update(i, a[i], 1, n);
  }
  int l, r, tp;
  while (m--) {
    cin >> tp;
    if (tp == 1) {
      arr.clear();
      cin >> l >> r;
      fnd(l, r, 1, n);
      for (int e : arr) {
        a[e] = p[a[e]];
        update(e, a[e], 1, n);
      }
    } else {
      cin >> l >> r;
      cout << get(l, r, 1, n) << '\n';
    }
  }
  return 0;
}
