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 MAXm = 1e6 + 6.66;
const long long MAXn = 3e5 + 6.66;
long long seg[4 * MAXn], lazy[4 * MAXn];
long long a[MAXn], Div[MAXm];
long long n, m, t, l, r;
void prep() {
  for (long long i = 1; i < MAXm; ++i)
    for (long long j = i; j < MAXm; j += i) Div[j]++;
}
void build(long long ind, long long l, long long r) {
  if (l + 1 == r) {
    seg[ind] = a[l];
    lazy[ind] = Div[seg[ind]] != seg[ind];
    return;
  }
  long long mid = l + (r - l) / 2;
  build(ind << 1, l, mid);
  build(ind << 1 | 1, mid, r);
  seg[ind] = seg[ind << 1] + seg[ind << 1 | 1];
  lazy[ind] = lazy[ind << 1] + lazy[ind << 1 | 1];
}
void update(long long ind, long long l, long long r, long long b, long long e) {
  if (l >= e || r <= b || !lazy[ind]) return;
  if (l + 1 == r) {
    seg[ind] = Div[seg[ind]];
    lazy[ind] = seg[ind] != Div[seg[ind]];
    return;
  }
  long long mid = l + (r - l) / 2;
  update(ind << 1, l, mid, b, e);
  update(ind << 1 | 1, mid, r, b, e);
  lazy[ind] = lazy[ind << 1] + lazy[ind << 1 | 1];
  seg[ind] = seg[ind << 1] + seg[ind << 1 | 1];
}
long long get(long long ind, long long l, long long r, long long b,
              long long e) {
  if (l >= e || r <= b) return 0;
  if (b <= l && r <= e) return seg[ind];
  long long mid = l + (r - l) / 2;
  return get(ind << 1, l, mid, b, e) + get(ind << 1 | 1, mid, r, b, e);
}
signed main() {
  ios::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  prep();
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  build(1, 0, MAXn);
  while (m--) {
    cin >> t >> l >> r;
    l--;
    t--;
    if (!t)
      update(1, 0, MAXn, l, r);
    else
      cout << get(1, 0, MAXn, l, r) << "\n";
  }
  return cout << endl, 0;
}