Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int d[1000001];
long long tree[4 * 1000001], mx[4 * 1000001], a[1000001];
void build(int node, int s, int e) {
  if (s == e)
    tree[node] = mx[node] = a[s];
  else {
    int m = (s + e) / 2;
    build(2 * node, s, m);
    build(2 * node + 1, m + 1, e);
    tree[node] = tree[2 * node] + tree[2 * node + 1];
    mx[node] = max(mx[2 * node], mx[2 * node + 1]);
  }
}
void update(int node, int s, int e, int l, int r) {
  if (mx[node] <= 2) return;
  if (s == e) {
    tree[node] = mx[node] = d[tree[node]];
    return;
  }
  int m = (s + e) / 2;
  if (l <= m) update(node * 2, s, m, l, r);
  if (r > m) update(node * 2 + 1, m + 1, e, l, r);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
  mx[node] = max(mx[2 * node], mx[2 * node + 1]);
}
long long query(int node, int s, int e, int l, int r) {
  if (l <= s and r >= e) return tree[node];
  long long m = (s + e) / 2;
  long long ans = 0;
  if (l <= m) ans += query(node * 2, s, m, l, r);
  if (r > m) ans += query(node * 2 + 1, m + 1, e, l, r);
  return ans;
}
int main() {
  for (int i = 1; i < 1000001; i++)
    for (int j = i; j < 1000001; j += i) d[j]++;
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  build(1, 1, n);
  while (m--) {
    int c, l, r;
    scanf("%d %d %d", &c, &l, &r);
    if (c == 1)
      update(1, 1, n, l, r);
    else
      cout << query(1, 1, n, l, r) << '\n';
  }
}