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 int n, m, a[1000100], t1[2 * 1000100], d[1000100];
long long int t[2 * 1000100];
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = a[l];
    t1[v] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(v + v, l, mid);
  build(v + v + 1, mid + 1, r);
  t[v] = t[v + v] + t[v + v + 1];
  t1[v] = max(t1[v + v], t1[v + v + 1]);
}
long long int get1(int v, int tl, int tr, int l, int r) {
  if (tl > r || tr < l) return 0;
  if (tl >= l && tr <= r) {
    return t1[v];
  }
  int tm = (tl + tr) / 2;
  return max(get1(v + v, tl, tm, l, r), get1(v + v + 1, tm + 1, tr, l, r));
}
void update(int v, int tl, int tr, int l, int r) {
  if (tl > r || tr < l) return;
  if (get1(1, 1, n, tl, tr) <= 2) {
    return;
  }
  if (tl == tr) {
    t[v] = t1[v] = d[t[v]];
    return;
  }
  int tm = (tl + tr) / 2;
  update(v + v, tl, tm, l, r);
  update(v + v + 1, tm + 1, tr, l, r);
  t[v] = t[v + v] + t[v + v + 1];
  t1[v] = max(t1[v + v], t1[v + v + 1]);
}
long long int get(int v, int tl, int tr, int l, int r) {
  if (tl > r || tr < l) return 0;
  if (tl >= l && tr <= r) {
    return t[v];
  }
  int tm = (tl + tr) / 2;
  return get(v + v, tl, tm, l, r) + get(v + v + 1, tm + 1, tr, l, r);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build(1, 1, n);
  for (int i = 1; i <= 1e6; ++i) {
    for (int j = i; j <= 1e6; j += i) {
      d[j]++;
    }
  }
  for (int i = 1, x, y, z; i <= m; ++i) {
    cin >> x >> y >> z;
    if (x == 1) {
      update(1, 1, n, y, z);
    } else {
      cout << get(1, 1, n, y, z) << "\n";
    }
  }
  return 0;
}