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;
struct seg {
  long long sum, mn, l, r;
};
seg a[1000030 * 4];
long long b[1000030], c[1000030], n, query;
void build(long long x, long long l, long long r) {
  a[x].l = l;
  a[x].r = r;
  if (l == r) {
    a[x].sum = a[x].mn = b[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
  a[x].sum = a[x * 2].sum + a[x * 2 + 1].sum;
  a[x].mn = max(a[x * 2].mn, a[x * 2 + 1].mn);
}
void update(long long x, long long l, long long r) {
  if (a[x].l > r || a[x].r < l) return;
  if (a[x].l >= l && a[x].r <= r && a[x].mn <= 2) return;
  if (a[x].l == a[x].r)
    b[a[x].l] = c[b[a[x].l]], a[x].sum = a[x].mn = b[a[x].l];
  else {
    update(x * 2, l, r);
    update(x * 2 + 1, l, r);
    a[x].sum = a[x * 2].sum + a[x * 2 + 1].sum;
    a[x].mn = max(a[x * 2].mn, a[x * 2 + 1].mn);
  }
}
long long qry(long long x, long long l, long long r) {
  if (a[x].l > r || a[x].r < l) return 0;
  if (a[x].l >= l && a[x].r <= r) return a[x].sum;
  return qry(x * 2, l, r) + qry(x * 2 + 1, l, r);
}
int main() {
  memset(c, 0, sizeof(c));
  for (long long i = 1; i < 1000030; i++)
    for (long long j = 1; i * j < 1000030; j++) c[i * j]++;
  scanf("%I64d%I64d", &n, &query);
  for (int i = 1; i <= n; i++) scanf("%I64d", &b[i]);
  build(1, 1, n);
  while (query--) {
    long long o, u, v;
    scanf("%I64d%I64d%I64d", &o, &u, &v);
    if (o == 1)
      update(1, u, v);
    else
      printf("%I64d\n", qry(1, u, v));
  }
  return 0;
}