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;
long long int st[18 * 100010], sum[3 * 100010];
int D[3 * 1000010], arr[3 * 100010];
void build(int root, int l, int r) {
  if (l == r) {
    st[root] = arr[l];
    return;
  }
  if (l > r) return;
  int left = root << 1, right = left | 1, mid = (l + r) >> 1;
  build(left, l, mid);
  build(right, mid + 1, r);
  st[root] = st[left] + st[right];
}
long long int query(int root, int l, int r, int rl, int rr) {
  if (rl <= l and rr >= r) return st[root];
  if (rl > r || rr < l) return 0;
  int left = root << 1, right = left | 1, mid = (l + r) >> 1;
  return query(left, l, mid, rl, rr) + query(right, mid + 1, r, rl, rr);
}
void update(int root, int l, int r, int rl) {
  if (rl <= l and rl >= r) {
    st[root] = arr[l];
    return;
  }
  if (rl > r || rl < l) return;
  int left = root << 1, right = left | 1, mid = (l + r) >> 1;
  update(left, l, mid, rl);
  update(right, mid + 1, r, rl);
  st[root] = st[left] + st[right];
}
int main() {
  int n, m, i, x, y;
  scanf("%d %d", &n, &m);
  set<int> s;
  for (int z = 0; z < n; z++) {
    scanf("%d", &arr[z]);
    if (arr[z] > 2) s.insert(z);
  }
  build(1, 0, n - 1);
  auto it = s.begin();
  for (int i = 1; i <= 1000000; i++)
    for (int j = i; j <= 1000000; j += i) D[j]++;
  auto it1 = it;
  bool k = true;
  while (m--) {
    scanf("%d", &i);
    scanf("%d %d", &x, &y);
    --x, --y;
    if ((int)s.size() == 0 and k) {
      k = false;
      sum[0] = arr[0];
      for (int z = 1; z < n; z++) sum[z] = arr[z] + sum[z - 1];
    }
    if (i == 2) {
      if ((int)s.size())
        printf("%lld\n", query(1, 0, n - 1, x, y));
      else
        printf("%lld\n", sum[y] - (x ? sum[x - 1] : 0));
    } else if (k) {
      it = s.lower_bound(x);
      while (it != s.end() and *it <= y) {
        arr[*it] = D[arr[*it]];
        update(1, 0, n - 1, *it);
        it1 = it;
        it1++;
        if (arr[*it] <= 2) s.erase(it);
        it = it1;
      }
    }
  }
  return 0;
}
