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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
long long arr[300005];
long long A[1000005];
long long tree[300005 + 5];
void fun() {
  for (long long i = 1; i <= 1e6 + 5; i++) {
    for (long long j = i; j < 1e6 + 5; j += i) A[j]++;
  }
}
long long query(long long idx) {
  long long sum = 0;
  while (idx > 0) {
    sum += tree[idx];
    idx -= idx & (-idx);
  }
  return sum;
}
void update(long long idx, long long x, long long n) {
  while (idx <= n) {
    tree[idx] += x;
    idx += idx & (-idx);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long tst = 1;
  for (long long tt = 1; tt <= tst; tt++) {
    fun();
    set<long long> s;
    long long n, q;
    cin >> n >> q;
    for (long long i = 1; i <= n; i++) {
      cin >> arr[i];
      update(i, arr[i], n);
      if (arr[i] > 2) s.insert(i);
    }
    while (q--) {
      long long l, r, t;
      cin >> t >> l >> r;
      if (t == 1) {
        while (1) {
          auto it = s.lower_bound(l);
          if (*it > r || it == s.end()) break;
          l = *it;
          update(l, A[arr[l]] - arr[l], n);
          arr[l] = A[arr[l]];
          if (arr[l] <= 2) s.erase(it);
          l++;
        }
      } else
        cout << query(r) - query(l - 1) << endl;
    }
  }
  return 0;
}