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 n, m, a[300005], itmax[1200005], d[1000005];
long long it[1200005];
void build_tree(int k, int l, int r) {
  if (l == r) {
    it[k] = a[l];
    itmax[k] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build_tree(k * 2, l, mid);
  build_tree(k * 2 + 1, mid + 1, r);
  it[k] = it[k * 2] + it[k * 2 + 1];
  itmax[k] = max(itmax[k * 2], itmax[k * 2 + 1]);
}
void update(int k, int l, int r, int L, int R) {
  if (r < L || R < l || L > R || l > r) return;
  if (itmax[k] <= 2) return;
  if (L <= l && r <= R && l == r) {
    itmax[k] = d[it[k]];
    it[k] = d[it[k]];
    return;
  }
  int mid = (l + r) / 2;
  update(2 * k, l, mid, L, R);
  update(2 * k + 1, mid + 1, r, L, R);
  itmax[k] = max(itmax[k * 2], itmax[k * 2 + 1]);
  it[k] = it[k * 2] + it[k * 2 + 1];
  return;
}
long long get(int k, int l, int r, int L, int R) {
  if (r < L || R < l || L > R || l > r) return 0;
  if (L <= l && r <= R) {
    return it[k];
  }
  int mid = (l + r) / 2;
  return get(k * 2, l, mid, L, R) + get(k * 2 + 1, mid + 1, r, L, R);
  it[k] = it[k * 2] + it[k * 2 + 1];
  itmax[k] = max(itmax[k * 2], itmax[k * 2 + 1]);
}
int main() {
  memset(d, 0, sizeof(d));
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build_tree(1, 1, n);
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j = j + i) {
      d[j]++;
    }
  }
  int type;
  int l, r;
  while (m--) {
    cin >> type >> l >> r;
    if (type == 1) {
      update(1, 1, n, l, r);
    } else {
      cout << get(1, 1, n, l, r) << "\n";
    }
  }
}