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;
int ar[2000005];
int divi[2000005];
int n, m;
long long sum[2000005];
set<int> idx;
void update(int node, int be, int en, int pos) {
  if (be > pos || en < pos || be > en) return;
  if (be == en) {
    sum[node] = ar[pos];
    return;
  }
  int mid = (be + en) / 2;
  update(node * 2, be, mid, pos);
  update(node * 2 + 1, mid + 1, en, pos);
  sum[node] = sum[node * 2] + sum[node * 2 + 1];
}
long long query(int node, int be, int en, int i, int j) {
  if (en < i || be > j || be > en) return 0;
  if (be >= i && en <= j) {
    return sum[node];
  }
  int mid = (be + en) / 2;
  return query(node * 2, be, mid, i, j) +
         query(node * 2 + 1, mid + 1, en, i, j);
}
void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ar[i]);
    if (ar[i] > 2) idx.insert(i);
    update(1, 1, n, i);
  }
  while (m--) {
    int ti, l, r;
    scanf("%d%d%d", &ti, &l, &r);
    if (ti == 1) {
      auto be = idx.lower_bound(l);
      auto en = idx.upper_bound(r);
      for (auto it = be; it != en; it++) {
        int pos = *it;
        ar[pos] = divi[ar[pos]];
        update(1, 1, n, pos);
      }
      for (auto it = be; it != en;) {
        auto tem = it;
        it++;
        int pos = *tem;
        if (ar[pos] <= 2) idx.erase(tem);
      }
    } else {
      printf("%lld\n", query(1, 1, n, l, r));
    }
  }
}
int main() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) divi[j]++;
  }
  int t = 1;
  while (t--) solve();
  return 0;
}