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 N = 312345;
const int M = 1123456;
long long v[N], divi[M], nd[M];
map<long long, long long> pri[M];
void prep() {
  for (int i = 2; i * i <= 1000000; i++) {
    if (divi[i] == 0) {
      for (int j = i; j <= 1000000; j += i) {
        divi[j] = i;
      }
    }
  }
  for (long long i = 2; i <= 1000000; i++) {
    if (divi[i] == 0) divi[i] = i;
  }
  nd[1] = 1;
  for (long long i = 2; i <= 1000000; i++) {
    long long aux = i;
    while (aux > 1) {
      pri[i][divi[aux]]++;
      aux /= divi[aux];
    }
    nd[i] = 1;
    for (auto x : pri[i]) {
      nd[i] *= x.second + 1;
    }
  }
}
int n, a, b;
long long bit[N];
long long query(int id) {
  long long sum = 0;
  while (id > 0) {
    sum += bit[id];
    id -= id & -id;
  }
  return sum;
}
void update(int id, long long val) {
  while (id <= N) {
    bit[id] += val;
    id += id & -id;
  }
}
int main(void) {
  prep();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
  for (int i = 1; i <= n; i++) update(i, v[i]);
  set<int> act;
  for (int i = 1; i <= n; i++) {
    if (v[i] > 2) act.emplace(i);
  }
  for (int i = 0; i < m; i++) {
    int type;
    cin >> type;
    scanf("%d%d", &a, &b);
    if (type == 1) {
      auto it = act.lower_bound(a);
      while (it != act.end()) {
        int val = *it;
        if (val > b) break;
        long long que = query(val) - query(val - 1);
        update(val, -que + nd[que]);
        if (query(val) - query(val - 1) <= 2)
          it = act.erase(it);
        else
          it++;
      }
    } else {
      printf("%lld\n", query(b) - query(a - 1));
    }
  }
  return 0;
}