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 cnt[1000005], a[300005], tree[1200000];
void update(long long v, long long l, long long r, long long pos,
            long long val) {
  if (l == r) {
    tree[v] = val;
    return;
  }
  long long mid = (l + r) / 2;
  if (pos <= mid) {
    update(v * 2, l, mid, pos, val);
  } else {
    update(v * 2 + 1, mid + 1, r, pos, val);
  }
  tree[v] = tree[v * 2] + tree[v * 2 + 1];
}
long long get_sum(long long v, long long l, long long r, long long nl,
                  long long nr) {
  if (nl > nr) {
    return 0;
  }
  if (l == nl && nr == r) {
    return tree[v];
  }
  long long mid = (l + r) / 2;
  return get_sum(v * 2, l, mid, nl, min(nr, mid)) +
         get_sum(v * 2 + 1, mid + 1, r, max(mid + 1, nl), nr);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, i, j;
  cin >> n >> m;
  set<long long> bad;
  set<long long>::iterator ii;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    update(1, 1, n, i, a[i]);
    if (a[i] > 2) {
      bad.insert(i);
    }
  }
  bad.insert(n + 1);
  for (i = 1; i <= 1000000; i++) {
    for (j = i; j <= 1000000; j += i) {
      cnt[j]++;
    }
  }
  while (m--) {
    long long t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      vector<long long> kick;
      ii = bad.lower_bound(l);
      while (*ii <= r) {
        long long cur = *ii;
        a[cur] = cnt[a[cur]];
        if (a[cur] == 2) {
          kick.push_back(cur);
        }
        update(1, 1, n, cur, a[cur]);
        ii++;
      }
      for (i = 0; i < kick.size(); i++) {
        long long cur = kick[i];
        bad.erase(cur);
      }
    } else {
      cout << get_sum(1, 1, n, l, r) << "\n";
    }
  }
  return 0;
}