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>
const int M = 300010, N = 1000000;
int n, m;
int d[N + 110];
int a[M], f[M];
template <typename T>
inline void read(T &x) {
  x = 0;
  char ch = getchar();
  int f = 1;
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
namespace tr {
long long tr[N];
inline void insert(int pos, int num) {
  for (; pos <= n; pos += pos & (-pos)) tr[pos] += num;
}
inline long long query(int pos) {
  long long ret = 0;
  for (; pos; pos -= pos & (-pos)) ret += tr[pos];
  return ret;
}
}  // namespace tr
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int main() {
  for (int i = 1; i <= N; i++)
    for (int j = i; j <= N; j += i) d[j]++;
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    tr::insert(i, a[i]);
    f[i] = i;
  }
  f[n + 1] = n + 1;
  while (m--) {
    int opt, l, r;
    read(opt);
    read(l);
    read(r);
    if (l > r) std::swap(l, r);
    if (opt == 1)
      for (int i = l; i <= r;) {
        tr::insert(i, d[a[i]] - a[i]);
        a[i] = d[a[i]];
        if (a[i] <= 2) f[i] = i + 1;
        if (i == find(i))
          i++;
        else
          i = f[i];
      }
    else
      printf("%lld\n", tr::query(r) - tr::query(l - 1));
  }
  return 0;
}