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 = 1000000;
const int MAXN = 300010;
int D[N + 1], a[MAXN], n, m;
long long dt[MAXN];
void add(int x, int v) {
  for (int i = x; i <= n; i += i & (-i)) dt[i] += v;
}
long long sum(int x) {
  long long res = 0;
  for (int i = x; i; i -= i & (-i)) {
    res += dt[i];
  }
  return res;
}
int main() {
  for (int i = 1; i <= N; ++i)
    for (int j = i; j <= N; j += i) D[j]++;
  scanf("%d%d", &n, &m);
  set<int> st;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > 2) st.insert(i);
    add(i, a[i]);
  }
  while (m--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      auto first = st.lower_bound(l);
      auto last = st.upper_bound(r);
      queue<int> tmp;
      while (first != last) {
        int i = *first;
        int x = D[a[i]];
        add(i, x - a[i]);
        a[i] = x;
        if (a[i] <= 2) tmp.push(i);
        first++;
      }
      while (!tmp.empty()) {
        int x = tmp.front();
        tmp.pop();
        st.erase(x);
      }
    } else {
      printf("%lld\n", sum(r) - sum(l - 1));
    }
  }
  return 0;
}
