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;
long long arr[300005], tree[2000009], tree2[2000009];
int cnt[1000010];
void build(long long node, long long l, long long r) {
  if (l > r) return;
  if (l == r) {
    tree[node] = arr[l];
    tree2[node] = (tree[node] != 2 && tree[node] != 1);
    return;
  }
  build(node * 2, l, (l + r) / 2);
  build(node * 2 + 1, (l + r) / 2 + 1, r);
  tree[node] = (tree[node * 2] + tree[node * 2 + 1]);
  tree2[node] = tree2[node * 2] | tree2[node * 2 + 1];
}
void update(long long node, long long l, long long r, long long i,
            long long j) {
  if (i > r || j < l || l > r) return;
  if (l == r) {
    tree[node] = cnt[tree[node]];
    tree2[node] = (tree[node] != 2 && tree[node] != 1);
    return;
  }
  if (tree2[node * 2]) update(node * 2, l, (l + r) / 2, i, j);
  if (tree2[node * 2 + 1]) update(node * 2 + 1, (l + r) / 2 + 1, r, i, j);
  tree[node] = (tree[node * 2] + tree[node * 2 + 1]);
  tree2[node] = tree2[node * 2] | tree2[node * 2 + 1];
}
long long query(long long node, long long l, long long r, long long i,
                long long j) {
  if (i > r || l > j || l > r) return 0;
  if (l >= i && r <= j) {
    return tree[node];
  }
  long long left = query(node * 2, l, (l + r) / 2, i, j);
  long long right = query(node * 2 + 1, (l + r) / 2 + 1, r, i, j);
  return (left + right);
}
int main() {
  long long i, j, k;
  long long n, m;
  for (i = 1; i <= 1000000; i++) {
    for (j = i; j <= 1000000; j += i) cnt[j]++;
  }
  scanf("%lld", &n);
  scanf("%lld", &m);
  for (i = 1; i <= n; i++) scanf("%lld", &arr[i]);
  build(1, 1, n);
  while (m--) {
    long long type;
    scanf("%lld", &type);
    if (type == 1) {
      scanf("%lld", &i);
      scanf("%lld", &j);
      update(1, 1, n, i, j);
    } else if (type == 2) {
      scanf("%lld", &i);
      scanf("%lld", &j);
      printf("%lld\n", query(1, 1, n, i, j));
    }
  }
  return 0;
}