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 node {
  bool fin;
  long long val;
  node *lc, *rc;
  node(int v) : fin(false), val(v), lc(nullptr), rc(nullptr){};
  void pull() { val = lc->val + rc->val; }
};
int n, m, arr[300001], D[1000001];
int cc(int k) {
  if (D[k] != -1) return D[k];
  int val = 0, j = 1;
  for (j = 1; j * j < k; j++)
    if (k % j == 0) val += 2;
  if (j * j == k) val++;
  return D[k] = val;
}
node *bld(int l, int r, node *p) {
  if (l == r) return new node(arr[l]);
  p = new node(0);
  int m = (l + r) / 2;
  p->lc = bld(l, m, p->lc), p->rc = bld(m + 1, r, p->rc);
  p->pull();
  return p;
}
long long qry(int ql, int qr, int l, int r, node *p) {
  if (ql <= l && r <= qr) return p->val;
  int m = (l + r) / 2;
  if (qr <= m) return qry(ql, qr, l, m, p->lc);
  if (ql > m) return qry(ql, qr, m + 1, r, p->rc);
  return qry(ql, qr, l, m, p->lc) + qry(ql, qr, m + 1, r, p->rc);
}
void mod(int ml, int mr, int l, int r, node *p) {
  if (p->fin) return;
  if (l == r) {
    p->val = cc(p->val);
    if (p->val <= 2) p->fin = true;
    return;
  }
  int m = (l + r) / 2;
  if (ml <= m) mod(ml, mr, l, m, p->lc);
  if (mr > m) mod(ml, mr, m + 1, r, p->rc);
  if (p->lc->fin && p->rc->fin) p->fin = true;
  p->pull();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 1; i <= 1000000; i++) D[i] = -1;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  node *rt = bld(1, n, rt);
  int a, b, c;
  while (m--) {
    cin >> a >> b >> c;
    if (a == 1)
      mod(b, c, 1, n, rt);
    else
      cout << qry(b, c, 1, n, rt) << '\n';
  }
  return 0;
}