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;
long long MOD = 1000000007;
long long bigmod(long long a, long long b) {
  if (b == 0) return 1;
  long long x = bigmod(a, b / 2);
  x = (x * x) % MOD;
  if (b % 2 == 1) x *= a;
  return x % MOD;
}
long long BIT[300005], a[300005], n, b[300005];
void update(int x, int delta) {
  for (; x <= n; x += x & -x) BIT[x] += delta;
}
long long query(int x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
long long sum(int l, int r) { return query(r) - query(l - 1); }
int cnt[1000005];
void sieve() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) cnt[j]++;
  }
}
int m;
int nxt[300005], pre[300005];
int main() {
  ios::sync_with_stdio(0);
  sieve();
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    b[i] = x;
    update(i, x);
  }
  for (int i = 0; i <= n; i++) {
    nxt[i] = i + 1;
    pre[i + 1] = i;
  }
  set<int> se;
  for (int i = 1; i <= n + 1; i++) se.insert(i);
  while (m--) {
    int t, l, r;
    cin >> t >> l >> r;
    if (t == 1) {
      set<int>::iterator it = se.lower_bound(l), temp;
      while (*it <= r) {
        int x = *it;
        int s = b[x];
        int c = cnt[s];
        update(x, c - s);
        temp = it;
        b[x] += (c - s);
        it++;
        if (b[x] == 1 || b[x] == 2) {
          se.erase(temp);
        }
      }
    } else {
      cout << sum(l, r) << endl;
    }
  }
  return 0;
}
