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 IT_MAX = 1 << 19;
long long indt[1100000];
void update(int p, long long v) {
  p += IT_MAX - 1;
  indt[p] = v;
  for (p /= 2; p; p /= 2) indt[p] = indt[2 * p] + indt[2 * p + 1];
}
long long getsum(int p1, int p2) {
  p1 += IT_MAX - 1;
  p2 += IT_MAX - 1;
  long long rv = 0;
  for (; p1 <= p2; p1 /= 2, p2 /= 2) {
    if (p1 % 2 == 1) rv = rv + indt[p1++];
    if (p2 % 2 == 0) rv = rv + indt[p2--];
  }
  return rv;
}
set<int> Sx;
int in[300050];
int D[1000050];
vector<int> Vl;
int main() {
  int N, M, i, j;
  scanf("%d %d", &N, &M);
  for (i = 1; i <= N; i++) scanf("%d", &in[i]);
  for (i = 1; i <= N; i++) update(i, in[i]);
  for (i = 1; i <= N; i++)
    if (in[i] >= 3) Sx.insert(i);
  for (i = 1; i <= 1000000; i++)
    for (j = i; j <= 1000000; j += i) D[j]++;
  while (M--) {
    int t1, t2, t3;
    scanf("%d %d %d", &t1, &t2, &t3);
    if (t1 == 1) {
      while (1) {
        set<int>::iterator IT = Sx.lower_bound(t2);
        if (IT == Sx.end() || *IT > t3) {
          break;
        }
        Vl.push_back(*IT);
        Sx.erase(IT);
      }
      for (int i = 0; i < Vl.size(); i++) {
        in[Vl[i]] = D[in[Vl[i]]];
        update(Vl[i], in[Vl[i]]);
        if (in[Vl[i]] >= 3) Sx.insert(Vl[i]);
      }
      Vl.clear();
    } else
      printf("%lld\n", getsum(t2, t3));
  }
  return 0;
}
