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;
const int N = 1e6 + 5;
int n, m, a[N], d[N];
struct BIT {
  long long c[N];
  void init() { memset(c, 0, sizeof c); }
  void update(int x, long long y) {
    for (; x < N; x += (x & (-x))) c[x] += y;
  }
  long long query(int x) {
    long long res = 0;
    for (; x; x -= (x & (-x))) res += c[x];
    return res;
  }
} bi;
struct DSU {
  int fa[N];
  void init() {
    for (int i = 0; i < N; i++) fa[i] = i;
  }
  int Find(int x) { return x == fa[x] ? x : fa[x] = Find(fa[x]); }
} ds;
int main() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) d[j]++;
  }
  bi.init();
  ds.init();
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    bi.update(i, a[i]);
  }
  while (m--) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) {
      for (int i = l; i <= r;) {
        bi.update(i, -a[i]);
        bi.update(i, d[a[i]]);
        a[i] = d[a[i]];
        if (a[i] <= 2) ds.fa[i] = i + 1;
        i = i == ds.Find(i) ? i + 1 : ds.Find(i);
      }
    } else
      printf("%lld\n", bi.query(r) - bi.query(l - 1));
  }
  return 0;
}
