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 V = int(1e6 + 2);
const int N = int(3e5 + 2);
int n, m, a[N];
int divisors[V];
long long aib[N];
set<int> good;
void update(int pos, int value) {
  for (; pos <= n; pos += pos & -pos) {
    aib[pos] += value;
  }
}
long long query(int pos) {
  long long Sol = 0;
  for (; pos > 0; pos -= pos & -pos) Sol += aib[pos];
  return Sol;
}
void Replace(int l, int r) {
  int curr = l - 1;
  while (!good.empty()) {
    auto it = good.upper_bound(curr);
    if (it == end(good) || *it > r) break;
    int pos = *it;
    curr = pos;
    update(pos, -a[pos] + divisors[a[pos]]);
    a[pos] = divisors[a[pos]];
    if (a[pos] <= 2) {
      good.erase(it);
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  for (int i = 1; i < V; ++i) {
    for (int j = i; j < V; j += i) {
      divisors[j]++;
    }
  }
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), update(i, a[i]);
  for (int i = 1; i <= n; ++i) good.insert(i);
  for (int Q = 1; Q <= m; ++Q) {
    int tip, l, r;
    scanf("%d %d %d", &tip, &l, &r);
    if (tip == 1) {
      Replace(l, r);
    } else {
      long long Sum = query(r) - query(l - 1);
      printf("%I64d\n", Sum);
    }
  }
}