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 long long N = 1e6 + 10;
long long n, q, d[N];
long long a[N << 2], m[N << 2], s[N << 2];
void init() {
  for (long long i = 1; i < N; i++) {
    for (long long j = i; j < N; j += i) {
      d[j]++;
    }
  }
}
void build(long long t, long long l, long long r) {
  if (l == r) {
    m[t] = s[t] = a[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(t << 1, l, mid);
  build(t << 1 | 1, mid + 1, r);
  s[t] = s[t << 1] + s[t << 1 | 1];
  m[t] = max(m[t << 1], m[t << 1 | 1]);
}
void update(long long dl, long long dr, long long t, long long l, long long r) {
  if (m[t] <= 2) return;
  if (l == r) {
    s[t] = m[t] = d[s[t]];
    return;
  }
  long long mid = (l + r) >> 1;
  if (dl <= mid) update(dl, dr, t << 1, l, mid);
  if (dr > mid) update(dl, dr, t << 1 | 1, mid + 1, r);
  s[t] = s[t << 1] + s[t << 1 | 1];
  m[t] = max(m[t << 1], m[t << 1 | 1]);
}
long long query(long long dl, long long dr, long long t, long long l,
                long long r) {
  long long res = 0;
  if (dl <= l && r <= dr) {
    return s[t];
  }
  long long mid = (l + r) >> 1;
  if (dl <= mid) res += query(dl, dr, t << 1, l, mid);
  if (dr > mid) res += query(dl, dr, t << 1 | 1, mid + 1, r);
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie();
  init();
  cin >> n >> q;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  while (q--) {
    long long f, l, r;
    cin >> f >> l >> r;
    if (f == 1) {
      update(l, r, 1, 1, n);
    } else if (f == 2) {
      cout << query(l, r, 1, 1, n) << endl;
      ;
    }
  }
  return 0;
}
