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;
const int N = 300005, A = 1000002;
int n, m, a[N], divs[A];
struct node {
  long long val;
  bool isAllOnes;
} tree[4 * N];
void build(int l, int r, int i) {
  if (l == r) {
    tree[i] = {a[l], a[l] == 1};
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, i << 1), build(mid + 1, r, (i << 1) | 1);
  tree[i].val = tree[i << 1].val + tree[(i << 1) | 1].val;
  tree[i].isAllOnes = tree[i << 1].isAllOnes && tree[(i << 1) | 1].isAllOnes;
}
void update(int l, int r, int i, int S, int E) {
  if (r < S || E < l || tree[i].isAllOnes) return;
  if (l == r) {
    tree[i].val = a[l] = divs[a[l]];
    tree[i].isAllOnes = (a[l] <= 2);
    return;
  }
  int mid = (l + r) >> 1;
  update(l, mid, i << 1, S, E), update(mid + 1, r, (i << 1) | 1, S, E);
  tree[i].val = tree[i << 1].val + tree[(i << 1) | 1].val;
  tree[i].isAllOnes = tree[i << 1].isAllOnes && tree[(i << 1) | 1].isAllOnes;
}
long long query(int l, int r, int i, int S, int E) {
  if (r < S || E < l) return 0;
  if (S <= l && r <= E) return tree[i].val;
  int mid = (l + r) >> 1;
  return query(l, mid, i << 1, S, E) + query(mid + 1, r, (i << 1) | 1, S, E);
}
void init() {
  for (int(i) = (1); (i) <= (int)(A - 1); (i)++)
    for (long long j = i; j < A; j += i) ++divs[j];
}
int main() {
  init();
  scanf("%d%d", &n, &m);
  for (int(i) = (1); (i) <= (int)(n); (i)++) scanf("%d", &a[i]);
  build(1, n, 1);
  while (m--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1)
      update(1, n, 1, l, r);
    else
      printf("%lld\n", query(1, n, 1, l, r));
  }
  return 0;
}
