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;
using ll = long long;
const int mxN = 3e5 + 1, mxV = 1e6 + 1;
ll n, q, spf[mxV], d[mxV], a[mxN];
struct node {
  ll sum, mx;
} t[mxN << 2];
node merge(node a, node b) { return node{a.sum + b.sum, max(a.mx, b.mx)}; }
void build(int l = 0, int r = n - 1, int v = 1) {
  if (l == r) {
    t[v] = node{a[l], a[l]};
    return;
  }
  int m = l + r >> 1;
  build(l, m, v << 1);
  build(m + 1, r, v << 1 | 1);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
ll query(int l, int r, int lo = 0, int hi = n - 1, int v = 1) {
  if (l > hi || r < lo) return 0;
  if (lo >= l && hi <= r) return t[v].sum;
  int m = lo + hi >> 1;
  return query(l, r, lo, m, v << 1) + query(l, r, m + 1, hi, v << 1 | 1);
}
void upd(int l, int r, int lo = 0, int hi = n - 1, int v = 1) {
  if (l > hi || r < lo || t[v].mx < 3) return;
  if (lo == hi) {
    ll val = t[v].sum;
    t[v] = node{d[val], d[val]};
    return;
  }
  int m = lo + hi >> 1;
  upd(l, r, lo, m, v << 1);
  upd(l, r, m + 1, hi, v << 1 | 1);
  t[v] = merge(t[v << 1], t[v << 1 | 1]);
}
void sieve() {
  for (int i = 1; i < mxV; ++i) spf[i] = (i & 1) ? i : 2;
  for (int i = 3; i * i < mxV; ++i)
    if (spf[i] == i)
      for (int j = i * i; j < mxV; j += i)
        if (spf[j] == j) spf[j] = i;
}
int get(int x) {
  map<int, int> pf;
  while (x > 1) ++pf[spf[x]], x /= spf[x];
  int ret = 1;
  for (auto p : pf) ret *= (p.second + 1);
  return ret;
}
int main() {
  sieve();
  for (int i = 1; i < mxV; ++i) d[i] = get(i);
  cin.tie(0)->sync_with_stdio(0);
  cin >> n >> q;
  for (int i = 0; i < n; ++i) cin >> a[i];
  build();
  while (q--) {
    int t, l, r;
    cin >> t >> l >> r, --l, --r;
    if (t & 1)
      upd(l, r);
    else
      cout << query(l, r) << '\n';
  }
}