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 input() {
  int res = 0;
  char c = ' ';
  while (c < '0') c = getchar();
  while (c >= '0') res = res * 10 + (c - '0'), c = getchar();
  return res;
}
const long long N = 3e5 + 1, M = 1e6 + 1;
long long f[N];
void modify(int i, long long val) {
  for (; i < N; i |= (i + 1)) f[i] += val;
}
long long get(int i) {
  long long res = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) res += f[i];
  return res;
}
long long a[N], lp[M], cnt[M];
long long D(long long x) {
  long long res = 1;
  vector<long long> pos;
  while (x != 1) {
    long long y = lp[x];
    pos.push_back(y);
    res /= (cnt[y] + 1);
    ++cnt[y];
    res *= (cnt[y] + 1);
    x /= y;
  }
  for (int y : pos) cnt[y] = 0;
  return res;
}
set<long long> nt;
vector<long long> pr;
int main() {
  for (long long i = 2; i < M; ++i) {
    if (!lp[i]) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (long long j = 0; j < pr.size(); ++j) {
      if (pr[j] > lp[i]) break;
      if (i * pr[j] >= M) break;
      lp[i * pr[j]] = pr[j];
    }
  }
  int n = input(), q = input();
  for (int i = 1; i <= n; ++i) {
    a[i] = input();
    modify(i, a[i]);
    if (a[i] > 2) nt.insert(i);
  }
  for (int i = 0; i < q; ++i) {
    int t = input(), l = input(), r = input();
    if (t == 1) {
      vector<int> buf;
      while (nt.size()) {
        auto it = nt.lower_bound(l);
        if (it == nt.end() || *it > r) break;
        long long j = *it;
        long long d = D(a[j]);
        nt.erase(it);
        if (d > 2) buf.push_back(j);
        modify(j, -a[j] + d);
        a[j] = d;
      }
      for (int j : buf) nt.insert(j);
    } else {
      cout << get(r) - get(l - 1) << endl;
    }
  }
}