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 = 3e5 + 10, M = 1e6 + 5;
inline long long read() {
  long long x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
long long n, m, a[N], d[M + 5], fa[N], tre[N];
void shai() {
  for (int i = 1; i <= M; ++i)
    for (int j = i; j <= M; j += i) ++d[j];
}
inline void add(int x, long long a) {
  for (; x <= n; x += x & -x) tre[x] += a;
}
inline long long query(int x) {
  long long res = 0;
  for (; x; x -= x & -x) {
    res += tre[x];
  }
  return res;
}
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
signed main() {
  shai();
  int op, l, r;
  n = read();
  m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    add(i, a[i]);
    fa[i] = i;
  }
  fa[n + 1] = n + 1;
  while (m--) {
    op = read();
    l = read();
    r = read();
    if (l > r) swap(l, r);
    if (op & 1) {
      for (int i = l; i <= r;) {
        add(i, d[a[i]] - a[i]);
        a[i] = d[a[i]];
        if (a[i] <= 2) fa[i] = i + 1;
        if (i == find(i))
          ++i;
        else
          i = fa[i];
      }
    } else
      printf("%lld\n", query(r) - query(l - 1));
  }
}