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;
int a[300005];
int D[1000005];
int org[300005];
long long tree[4 * 300005];
void build(int node, int s, int e) {
  if (s == e) {
    tree[node] = a[s];
    return;
  }
  build(2 * node, s, (s + e) / 2);
  build(2 * node + 1, (s + e) / 2 + 1, e);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
void update(int node, int s, int e, int idx, int val) {
  if (s > idx || e < idx) return;
  if (s == e) {
    tree[node] = a[idx] = val;
    return;
  }
  update(node * 2, s, (s + e) / 2, idx, val);
  update(node * 2 + 1, (s + e) / 2 + 1, e, idx, val);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
long long sum(int node, int s, int e, int l, int r) {
  if (s > r || e < l) return 0;
  if (s >= l && e <= r) return tree[node];
  long long a = sum(node * 2, s, (s + e) / 2, l, r);
  long long b = sum(node * 2 + 1, (s + e) / 2 + 1, e, l, r);
  return a + b;
}
int main() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) D[j]++;
  }
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  build(1, 1, n);
  set<int> st;
  for (int i = 1; i <= n; i++) st.insert(i);
  while (m--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (x == 2) {
      printf("%I64d\n", sum(1, 1, n, y, z));
    } else {
      set<int>::iterator it = st.lower_bound(y);
      vector<int> r;
      for (; it != st.end() && *it <= z; it++) {
        update(1, 1, n, *it, D[a[*it]]);
        if (a[*it] == 1 || a[*it] == 2) r.push_back(*it);
      }
      for (int i = 0; i < r.size(); i++) st.erase(r[i]);
    }
  }
  return 0;
}
