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;
const int N = 300010;
const int MX = 1000010;
int n, m, t, l, r, a[N];
int d[MX];
long long s[N << 2], mx[N << 2];
void build(int x, int l, int r) {
  if (l == r) {
    s[x] = a[l];
    mx[x] = a[l];
    return;
  }
  int m = (l + r) >> 1;
  build(2 * x, l, m);
  build(2 * x + 1, m + 1, r);
  s[x] = s[2 * x] + s[2 * x + 1];
  mx[x] = max(mx[2 * x], mx[2 * x + 1]);
}
long long sum(int x, int l, int r, int lq, int rq) {
  if (lq <= l && r <= rq) return s[x];
  if (lq > r || rq < l) return 0;
  int m = (l + r) >> 1;
  return sum(2 * x, l, m, lq, rq) + sum(2 * x + 1, m + 1, r, lq, rq);
}
void upd(int x, int l, int r, int lq, int rq) {
  if (lq > r || rq < l) return;
  if (mx[x] < 3) return;
  if (l == r) {
    s[x] = d[s[x]];
    mx[x] = d[mx[x]];
    return;
  }
  int m = (l + r) >> 1;
  upd(2 * x, l, m, lq, rq);
  upd(2 * x + 1, m + 1, r, lq, rq);
  s[x] = s[2 * x] + s[2 * x + 1];
  mx[x] = max(mx[2 * x], mx[2 * x + 1]);
}
int main() {
  scanf("%d%d", &n, &m);
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    mx = max(mx, a[i]);
  }
  for (int i = 1; i <= mx; i++)
    for (int j = i; j <= mx; j += i) d[j]++;
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      upd(1, 1, n, l, r);
    } else {
      printf("%lld\n", sum(1, 1, n, l, r));
    }
  }
}