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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
long long A[1000005];
void fun() {
  for (long long i = 1; i <= 1e6 + 5; i++) {
    for (long long j = i; j < 1e6 + 5; j += i) A[j]++;
  }
}
long long segment[10000];
long long preprocess(long long inpt[], long long n) {
  long long current_segment = -1;
  long long segment_size = sqrt(n);
  for (long long i = 0; i < n; i++) {
    if (i % segment_size == 0) {
      current_segment++;
    }
    segment[current_segment] += inpt[i];
  }
  return segment_size;
}
long long query(long long inpt[], long long segment_size, long long l,
                long long r) {
  long long sum = 0;
  while (l < r && l % segment_size != 0) sum += inpt[l], l++;
  while (l + segment_size <= r) {
    sum += segment[l / segment_size];
    l += segment_size;
  }
  while (l <= r) sum += inpt[l], l++;
  return sum;
}
void update(long long inpt[], long long segment_size, long long i,
            long long val) {
  long long segment_no = i / segment_size;
  segment[segment_no] -= inpt[i];
  segment[segment_no] += val;
  inpt[i] = val;
}
int main() {
  long long tst = 1;
  for (long long tt = 1; tt <= tst; tt++) {
    fun();
    set<long long> s;
    long long n, q;
    scanf("%lld %lld", &n, &q);
    long long arr[n + 5];
    long long sz = sqrt(n);
    for (long long i = 0; i < n; i++) {
      scanf("%lld", &arr[i]);
      if (arr[i] > 2) s.insert(i);
    }
    preprocess(arr, n);
    while (q--) {
      long long t, l, r;
      scanf("%lld %lld %lld", &t, &l, &r);
      l--;
      r--;
      if (t == 1) {
        while (1) {
          auto it = s.lower_bound(l);
          if (*it > r || it == s.end()) break;
          l = *it;
          update(arr, sz, l, A[arr[l]]);
          if (arr[l] <= 2) s.erase(it);
          l++;
        }
      } else
        printf("%lld\n", query(arr, sz, l, r));
    }
  }
  return 0;
}