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;
int n, m, t, l, r, nums[300300], D[1001000];
int sum[550];
set<int> lft[550];
int main() {
  for (int i = 1; i <= 1000000; i++)
    for (int j = i; j <= 1000000; j += i) D[j]++;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &nums[i]);
    sum[i / 550] += nums[i];
    lft[i / 550].insert(i);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d %d %d", &t, &l, &r);
    l--;
    r--;
    if (t == 1) {
      if (l / 550 == r / 550) {
        vector<int> rem;
        for (int v : lft[l / 550]) {
          if (v < l || v > r) continue;
          if (nums[v] == D[nums[v]])
            rem.push_back(v);
          else {
            sum[l / 550] -= nums[v] - D[nums[v]];
            nums[v] = D[nums[v]];
          }
        }
        for (int v : rem) lft[l / 550].erase(v);
      } else {
        vector<int> rem;
        for (int v : lft[l / 550]) {
          if (v < l) continue;
          if (nums[v] == D[nums[v]])
            rem.push_back(v);
          else {
            sum[l / 550] -= nums[v] - D[nums[v]];
            nums[v] = D[nums[v]];
          }
        }
        for (int v : rem) lft[l / 550].erase(v);
        rem.clear();
        for (int j = l / 550 + 1; j < r / 550; j++) {
          for (int v : lft[j]) {
            if (nums[v] == D[nums[v]])
              rem.push_back(v);
            else {
              sum[j] -= nums[v] - D[nums[v]];
              nums[v] = D[nums[v]];
            }
          }
          for (int v : rem) lft[j].erase(v);
          rem.clear();
        }
        for (int v : lft[r / 550]) {
          if (v > r) continue;
          if (nums[v] == D[nums[v]])
            rem.push_back(v);
          else {
            sum[r / 550] -= nums[v] - D[nums[v]];
            nums[v] = D[nums[v]];
          }
        }
        for (int v : rem) lft[r / 550].erase(v);
        rem.clear();
      }
    } else {
      long long ans = 0;
      if (l / 550 == r / 550) {
        for (int j = l; j <= r; j++) ans += nums[j];
      } else {
        for (int j = l; j < (l / 550 + 1) * 550; j++) ans += nums[j];
        for (int j = l / 550 + 1; j < r / 550; j++) ans += sum[j];
        for (int j = r / 550 * 550; j <= r; j++) ans += nums[j];
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
