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;
const double PI = acos(-1.0);
template <typename T, typename U>
inline void amin(T& x, U y) {
  if (x > y) x = y;
}
template <typename T, typename U>
inline void amax(T& x, U y) {
  if (x < y) x = y;
}
int pre[1000005] = {0};
template <typename T>
struct bit {
  int n;
  vector<T> fen;
  bit() : n(0) {}
  bit(int _n) : n(_n), fen(_n + 10) {}
  void add(int p, T val) {
    for (; p <= n; p += p & -p) fen[p] += val;
  }
  T sum(int p) {
    T ans = 0;
    for (; p > 0; p -= p & -p) ans += fen[p];
    return ans;
  }
};
void prepro() {
  for (auto i = (1); i <= (1000005 - 5); i++)
    for (int j = i; j < 1000005; j += i) pre[j]++;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(17);
  int n, m;
  prepro();
  cin >> n >> m;
  bit<long long> t(n + 2);
  vector<int> a(n);
  set<int> q;
  for (auto i = (0); i <= ((n)-1); i++) {
    cin >> a[i];
    t.add(i + 1, a[i]);
    if (a[i] > 2) q.insert(i);
  }
  for (auto i__ = (0); i__ <= ((m)-1); i__++) {
    int id, l, r;
    cin >> id >> l >> r;
    if (id == 1) {
      vector<int> er;
      auto it = q.lower_bound(l - 1);
      for (; it != q.end() && *it < r; it++) {
        int i = *it;
        t.add(i + 1, pre[a[i]] - a[i]);
        a[i] = pre[a[i]];
        if (a[i] <= 2) er.push_back(i);
      }
      for (auto e : er) q.erase(e);
    } else
      cout << t.sum(r) - t.sum(l - 1) << "\n";
  };
  return 0;
}
