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;
const long long mod = 1e9 + 7;
const long long N = 3e5 + 5;
const long long A = 1e6 + 5;
const long double pi = acos(-1);
long long divs[A];
long long n, m, a[N];
void prepare() {
  for (long long i = 1; i < A; ++i)
    for (long long j = i; j < A; j += i) ++divs[j];
}
struct _node {
  long long sm;
  long long mx;
} node[N << 2];
void build(long long i, long long l, long long r) {
  if (l > r) return;
  if (l == r) {
    node[i].sm = node[i].mx = a[l];
    return;
  }
  long long mid = l + r >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
  node[i].sm = node[i << 1].sm + node[i << 1 | 1].sm;
  node[i].mx = max(node[i << 1].mx, node[i << 1 | 1].mx);
}
void update(long long i, long long l, long long r, long long L, long long R) {
  if (l > r || L > r || R < l) return;
  if (node[i].mx <= 2) return;
  if (l == r) {
    node[i].sm = node[i].mx = a[l] = divs[a[l]];
    return;
  }
  long long mid = l + r >> 1;
  update(i << 1, l, mid, L, R);
  update(i << 1 | 1, mid + 1, r, L, R);
  node[i].sm = node[i << 1].sm + node[i << 1 | 1].sm;
  node[i].mx = max(node[i << 1].mx, node[i << 1 | 1].mx);
}
long long get(long long i, long long l, long long r, long long L, long long R) {
  if (l > r || L > r || R < l) return 0;
  if (L <= l && R >= r) return node[i].sm;
  long long mid = l + r >> 1;
  return get(i << 1, l, mid, L, R) + get(i << 1 | 1, mid + 1, r, L, R);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  prepare();
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  build(1, 1, n);
  while (m--) {
    long long type, l, r;
    cin >> type >> l >> r;
    if (type == 1)
      update(1, 1, n, l, r);
    else
      cout << get(1, 1, n, l, r) << '\n';
  }
}