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;
const long long N = 1e6 + 6;
const long long M = 3e5 + 5;
long long cnt[N], a[M], n, m, t[2 * M];
set<long long> second;
set<long long>::iterator it;
void build_tree() {
  for (int i = n - 1; i > 0; i--) t[i] = t[2 * i] + t[2 * i + 1];
}
long long sum(long long l, long long r) {
  l += n - 1, r += n;
  long long ans = 0;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) ans += t[l++];
    if (r & 1) ans += t[--r];
  }
  return ans;
}
void update_val(long long x) {
  a[x] = cnt[a[x]];
  if (a[x] <= 2) second.erase(second.find(x));
  x--;
  x += n;
  t[x] = cnt[t[x]];
  while (x > 1) {
    t[x >> 1] = t[x] + t[x ^ 1];
    x >>= 1;
  }
}
void update(long long l, long long r) {
  while (1) {
    it = second.lower_bound(l);
    if (it == second.end()) break;
    if (*it > r) break;
    l = *it;
    update_val(l);
    l++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  memset(cnt, 0, sizeof(cnt));
  for (int i = 1; i < N; i++) {
    for (int j = 1; j < N; j++) {
      if (i * j >= N) break;
      cnt[i * j]++;
    }
  }
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] > 2) second.insert(i);
  }
  for (int i = 1; i <= n; i++) {
    t[i + n - 1] = a[i];
  }
  build_tree();
  while (m--) {
    long long type, l, r;
    cin >> type >> l >> r;
    if (type == 1) {
      update(l, r);
    } else {
      cout << sum(l, r) << endl;
    }
  }
}