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 int MAXN = 3e5;
const int MAXA = 1e6;
int n, m, type, l, r, a[MAXN + 1], d[MAXA + 1];
long long it[MAXN * 5];
bool fr[MAXN * 5];
void init(const int& lef = 1, const int& rig = n, const int& k = 1) {
  if (lef == rig) {
    it[k] = a[lef];
    return;
  }
  int mid = (lef + rig) >> 1;
  init(lef, mid, k << 1);
  init(mid + 1, rig, k << 1 | 1);
  it[k] = it[k << 1] + it[k << 1 | 1];
}
long long get(const int& u, const int& v, const int& lef = 1,
              const int& rig = n, const int& k = 1) {
  if (u == lef && v == rig) return it[k];
  int mid = (lef + rig) >> 1;
  if (v <= mid)
    return get(u, v, lef, mid, k << 1);
  else if (u > mid)
    return get(u, v, mid + 1, rig, k << 1 | 1);
  else
    return get(u, mid, lef, mid, k << 1) +
           get(mid + 1, v, mid + 1, rig, k << 1 | 1);
}
void down(const int& lef, const int& rig, const int& k) {
  if (fr[k]) return;
  if (lef == rig) {
    it[k] = d[it[k]];
    fr[k] = (it[k] == d[it[k]]);
    return;
  }
  int mid = (lef + rig) >> 1;
  down(lef, mid, k << 1);
  down(mid + 1, rig, k << 1 | 1);
  it[k] = it[k << 1] + it[k << 1 | 1];
  fr[k] = fr[k << 1] && fr[k << 1 | 1];
}
void up(const int& u, const int& v, const int& lef = 1, const int& rig = n,
        const int& k = 1) {
  if (u == lef && v == rig) {
    down(lef, rig, k);
    return;
  }
  int mid = (lef + rig) >> 1;
  if (v <= mid)
    up(u, v, lef, mid, k << 1);
  else if (u > mid)
    up(u, v, mid + 1, rig, k << 1 | 1);
  else {
    up(u, mid, lef, mid, k << 1);
    up(mid + 1, v, mid + 1, rig, k << 1 | 1);
  }
  it[k] = it[k << 1] + it[k << 1 | 1];
}
int main() {
  for (int i = 1; i <= MAXA; ++i)
    for (int j = 1; i * j <= MAXA; ++j) ++d[i * j];
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  init();
  while (m--) {
    scanf("%d %d %d", &type, &l, &r);
    if (type == 1)
      up(l, r);
    else
      printf("%lld\n", get(l, r));
  }
  return 0;
}