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;
void solve();
const int N = 1e6 + 2;
long long t[4 * N];
int a[N], D[N];
int n, m;
void upd(int node, int l, int r, int x) {
  t[node] = x;
  a[l] = x;
  return;
}
void build(int node = 1, int tl = 0, int tr = n - 1) {
  if (tl > tr) {
    return;
  }
  if (tl == tr) {
    t[node] = a[tl];
    return;
  }
  int mid = (tl + tr) / 2;
  build(2 * node, tl, mid);
  build(2 * node + 1, mid + 1, tr);
  t[node] = t[2 * node] + t[2 * node + 1];
}
void updateRange(int l, int r, int val, int node = 1, int tl = 0,
                 int tr = n - 1) {
  if (r < l or l > tr or tl > r or tl > tr) {
    return;
  }
  if (l <= tl and tr <= r) {
    upd(node, tl, tr, val);
    return;
  }
  int mid = (tl + tr) / 2;
  updateRange(l, r, val, 2 * node, tl, mid);
  updateRange(l, r, val, 2 * node + 1, mid + 1, tr);
  t[node] = t[2 * node] + t[2 * node + 1];
}
long long queryRange(int l, int r, int node = 1, int tl = 0, int tr = n - 1) {
  if (r < l or tr < tl or l > tr or tl > r) {
    return 0;
  }
  if (l <= tl and tr <= r) {
    return t[node];
  }
  int mid = (tl + tr) / 2;
  return (queryRange(l, r, 2 * node, tl, mid) +
          queryRange(l, r, 2 * node + 1, mid + 1, tr));
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  solve();
  cerr << "Time taken: " << ((clock() * 1000) / CLOCKS_PER_SEC) << "ms\n";
}
void solve() {
  memset(D, 0, sizeof(D));
  for (int i = 1; i < N; ++i) {
    for (int j = i; j < N; j += i) {
      D[j]++;
    }
  }
  cin >> n >> m;
  set<int> s;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    s.insert(i);
  }
  build();
  std::vector<int> v;
  for (int i = 0; i < m; ++i) {
    int T, l, r;
    cin >> T >> l >> r;
    l--, r--;
    if (T == 1) {
      auto it = s.lower_bound(l);
      while (it != s.end() and *it <= r) {
        if (a[*it] <= 2) {
          v.push_back(*it);
        } else {
          updateRange(*it, *it, D[a[*it]]);
        }
        ++it;
      }
      while (((int)v.size())) {
        s.erase(v.back());
        v.pop_back();
      }
    } else {
      cout << queryRange(l, r) << '\n';
    }
  }
}