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;
void fn() {}
inline int Rand() { return (rand() << 16) | rand(); }
const double eps = 1e-9, Pi = acos(-1);
const int N = 3e5 + 5, second = 2e3 + 5;
const int MOD = 1e9 + 7, inf = 21e8;
const long long INF = 9e18;
long long t[N * 4], g[N * 4], a[N * 2], was[N * 5];
void build(int v, int l, int r) {
  if (l == r) {
    t[v] = a[l], g[v] = (a[l] != was[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];
  g[v] = g[v + v] + g[v + v + 1];
}
void upd(int v, int l, int r, int ql, int qr) {
  if (ql > r || qr < l || g[v] == 0) return;
  if (l == r) {
    g[v] = (t[v] != was[t[v]]);
    t[v] = was[t[v]];
    return;
  }
  int mid = (l + r) / 2;
  upd(v + v, l, mid, ql, qr);
  upd(v + v + 1, mid + 1, r, ql, qr);
  t[v] = t[v + v] + t[v + v + 1];
  g[v] = g[v + v] + g[v + v + 1];
}
long long sum(int v, int l, int r, int ql, int qr) {
  if (ql > r || qr < l) return 0;
  if (ql <= l && r <= qr) {
    return t[v] * 1ll;
  }
  int mid = (l + r) / 2;
  return sum(v + v, l, mid, ql, qr) + sum(v + v + 1, mid + 1, r, ql, qr);
}
int main() {
  fn();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= 1e6; i++) {
    for (int j = i; j <= 1e6; j += i) {
      was[j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int x, l, r;
    scanf("%d %d %d", &x, &l, &r);
    if (x == 1)
      upd(1, 1, n, l, r);
    else
      printf("%I64d\n", sum(1, 1, n, l, r));
  }
  return 0;
}
