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;
int n, m, a[3 * 100010], d[12 * 100000 + 10];
long long t[12 * 100000 + 10];
int us[1000000 + 10];
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = a[l];
    d[v] = a[l];
  } else {
    int mid = (r + l) >> 1;
    build(2 * v, l, mid);
    build(2 * v + 1, mid + 1, r);
    t[v] = t[2 * v] + t[2 * v + 1];
    d[v] = max(d[2 * v], d[2 * v + 1]);
  }
}
void upd(int v, int l, int r, int L, int R) {
  if (L > R) return;
  if (d[v] < 3) return;
  if (l == r) {
    t[v] = us[t[v]];
    d[v] = us[d[v]];
  } else {
    int mid = (r + l) >> 1;
    if (d[2 * v] >= 3) upd(2 * v, l, mid, L, min(mid, R));
    if (d[2 * v + 1] >= 3) upd(2 * v + 1, mid + 1, r, max(mid + 1, L), R);
    t[v] = t[2 * v] + t[2 * v + 1];
    d[v] = max(d[2 * v], d[2 * v + 1]);
  }
}
long long get_sum(int v, int l, int r, int L, int R) {
  if (L > R) return 0;
  if (l == L && r == R) {
    return t[v];
  }
  int mid = (r + l) >> 1;
  return get_sum(2 * v, l, mid, L, min(R, mid)) +
         get_sum(2 * v + 1, mid + 1, r, max(L, mid + 1), R);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  us[1] = 1;
  for (int i = 2; i <= 1000000; i++) {
    int cnt = 1;
    int t = i;
    for (int j = 2; j * j <= t; j++) {
      if (t % j == 0) {
        int k = 0;
        while (t % j == 0) {
          t /= j;
          k++;
        }
        cnt *= (k + 1);
      }
    }
    if (t > 1) {
      cnt *= 2;
    }
    us[i] = cnt;
  }
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int tp, l, r;
    cin >> tp >> l >> r;
    if (tp == 1) {
      upd(1, 1, n, l, r);
    } else {
      cout << get_sum(1, 1, n, l, r) << '\n';
    }
  }
}