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 N = 3e5 + 5, MAXN = 1e6 + 6;
long long n, q, a[N], t[4 * N], tmax[4 * N], D[MAXN];
void sieve() {
  for (long long i = 1; i < MAXN; ++i) {
    for (long long j = i; j < MAXN; j += i) {
      D[j]++;
    }
  }
}
void build(long long v, long long tl, long long tr) {
  if (tl == tr) {
    t[v] = tmax[v] = a[tl];
    return;
  }
  long long tm = (tl + tr) >> 1LL;
  build(2 * v, tl, tm);
  build(2 * v + 1, tm + 1, tr);
  t[v] = t[2 * v] + t[2 * v + 1];
  tmax[v] = max(tmax[2 * v], tmax[2 * v + 1]);
}
long long solve(long long v, long long tl, long long tr, long long l,
                long long r) {
  if (l > r) return 0;
  if (tl == l and tr == r) return t[v];
  long long tm = (tl + tr) >> 1LL;
  long long left = solve(2 * v, tl, tm, l, min(r, tm));
  long long right = solve(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
  return (left + right);
}
void update(long long v, long long tl, long long tr, long long l, long long r) {
  if (l > r || tmax[v] <= 2) return;
  if (tl == tr) {
    t[v] = a[tl] = tmax[v] = D[a[tl]];
    return;
  }
  long long tm = (tl + tr) >> 1LL;
  update(2 * v, tl, tm, l, min(r, tm));
  update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
  t[v] = t[2 * v] + t[2 * v + 1];
  tmax[v] = max(tmax[2 * v], tmax[2 * v + 1]);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sieve();
  cin >> n >> q;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  build(1, 1, n);
  while (q--) {
    long long k, l, r;
    cin >> k >> l >> r;
    if (k == 1) {
      update(1, 1, n, l, r);
    } else {
      cout << solve(1, 1, n, l, r) << "\n";
    }
  }
}