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
#include <bits/stdc++.h>
using namespace std;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const long double pi = acos(-1);
struct BinaryIndexedTree {
  int N;
  vector<long long> v;
  BinaryIndexedTree(int N) : N(N), v(N + 1) {}
  long long sum(int i) {
    long long s = 0;
    for (; i > 0; s += v[i], i -= i & -i)
      ;
    return s;
  }
  void add(int i, long long x) {
    for (; i <= N; v[i] += x, i += i & -i)
      ;
  }
};
int N, M;
int A[300100];
int D[1001001];
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout << fixed;
  cout.tie(0);
  cin.tie(0);
  cin >> N >> M;
  BinaryIndexedTree bit(N);
  D[1] = 1;
  D[2] = 2;
  set<int> s;
  for (int i = 1; i <= N; i++) {
    cin >> A[i];
    if (A[i] >= 3) {
      s.insert(i);
    }
    bit.add(i, A[i]);
    int cur = A[i];
    while (!D[cur]) {
      int& cnt = D[cur];
      for (int d = 1; d * d <= cur; d++) {
        if (cur % d == 0) {
          cnt++;
          if (d * d < cur) {
            cnt++;
          }
        }
      }
      if (D[cnt] == 0) {
        cur = cnt;
      } else {
        break;
      }
    }
  }
  for (int query = 1; query <= M; query++) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      auto it = s.lower_bound(l);
      vector<int> rem;
      while (it != s.end() && *it <= r) {
        int i = *it;
        bit.add(i, D[A[i]] - A[i]);
        A[i] = D[A[i]];
        if (A[i] <= 2) {
          rem.push_back(i);
        }
        it++;
      }
      for (int x : rem) {
        s.erase(x);
      }
    } else {
      cout << bit.sum(r) - bit.sum(l - 1) << '\n';
    }
  }
}
