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 nax = 1e6 + 11;
const int dope = 3e5 + 11;
int D[nax];
long long tree[dope];
void ini() {
  int i, j;
  for (i = 1; i < nax; ++i)
    for (j = i; j < nax; j += i) D[j]++;
}
void update(int idx, int val) {
  if (idx < 0 || idx >= dope) return;
  for (idx++; idx < dope; idx += idx & -idx) tree[idx] += val;
}
long long query(int idx) {
  if (idx < -1 || idx >= dope) return 0LL;
  long long ret = 0;
  for (idx++; idx; idx -= idx & -idx) ret += tree[idx];
  return ret;
}
long long query(int L, int R) { return query(R) - query(L - 1); }
int main() {
  ini();
  int N, M, i, x, y;
  scanf("%d%d", &N, &M);
  vector<int> A(N);
  set<int> st;
  for (i = 0; i < N; ++i) {
    scanf("%d", &A[i]);
    update(i, A[i]);
    st.insert(i);
  }
  st.insert(-1);
  st.insert(N);
  for (; M; --M) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    --l;
    --r;
    if (t == 1) {
      set<int> toRemove;
      auto it = st.lower_bound(l);
      for (; *it <= r; ++it) {
        update(*it, -A[*it]);
        A[*it] = D[A[*it]];
        update(*it, A[*it]);
        if (A[*it] <= 2) toRemove.insert(*it);
      }
      for (auto v : toRemove) st.erase(v);
    } else {
      printf("%lld\n", query(l, r));
    }
  }
  return 0;
}
