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;
using ll = long long;
const int N = 3e5 + 5;
const int M = 1e6 + 5;
int a[N], c[M];
ll sg[4 * N];
int ok[4 * N];
vector<int> primes;
void build(int node, int l, int r) {
  if (l == r) {
    sg[node] = a[l];
    if (a[l] == 1 || a[l] == 2) ok[node] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(2 * node, l, mid);
  build(2 * node + 1, mid + 1, r);
  sg[node] = sg[2 * node] + sg[2 * node + 1];
  ok[node] = ok[2 * node] && ok[2 * node + 1];
}
ll que(int node, int l, int r, int le, int re) {
  if (r < le || l > re) return 0;
  if (l >= le && r <= re) return sg[node];
  int mid = (l + r) / 2;
  ll q1 = que(2 * node, l, mid, le, re);
  ll q2 = que(2 * node + 1, mid + 1, r, le, re);
  return q1 + q2;
}
void update(int node, int l, int r, int le, int re) {
  if (r < le || l > re) return;
  if (ok[node]) return;
  if (l == r) {
    sg[node] = c[sg[node]];
    if (sg[node] == 1 || sg[node] == 2) {
      ok[node] = true;
    }
    return;
  }
  int mid = (l + r) / 2;
  update(2 * node, l, mid, le, re);
  update(2 * node + 1, mid + 1, r, le, re);
  sg[node] = sg[2 * node] + sg[2 * node + 1];
  ok[node] = ok[2 * node] && ok[2 * node + 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  c[1] = 1;
  for (int i = 2; i < M; i++) {
    if (c[i] == 0) {
      primes.push_back(i);
      c[i] = 2;
    }
    for (int j = 0; j < primes.size() && 1ll * primes[j] * i < M; j++) {
      if (i % primes[j] == 0) {
        c[i * primes[j]] = 2 * c[i] - c[i / primes[j]];
        break;
      } else {
        c[i * primes[j]] = 2 * c[i];
      }
    }
  }
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  build(1, 0, n - 1);
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r;
    l--;
    r--;
    if (t == 1) {
      update(1, 0, n - 1, l, r);
    } else {
      cout << que(1, 0, n - 1, l, r) << "\n";
    }
  }
  return 0;
}