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 N = 3e5;
const long long M = 1e6 + 1;
struct node {
  int mx;
  long long sum;
  node() {
    mx = 0;
    sum = 0LL;
  }
};
node t[4 * N];
int arr[N];
int sp[M];
int d[M];
node join(const node &a, const node &b) {
  node res;
  res.mx = max(a.mx, b.mx);
  res.sum = a.sum + b.sum;
  return res;
}
void build(int i, int l, int r) {
  if (l == r) {
    t[i].mx = t[i].sum = arr[l];
    return;
  }
  int m = (l + r) >> 1;
  build(i << 1, l, m);
  build(i << 1 | 1, m + 1, r);
  t[i] = join(t[i << 1], t[i << 1 | 1]);
}
void update(int i, int l, int r, int x, int y) {
  if (l == r) {
    t[i].mx = t[i].sum = d[t[i].mx];
    return;
  }
  int m = (l + r) >> 1;
  int lft = i << 1;
  int rgt = lft | 1;
  if (x <= m && t[lft].mx > 2) {
    update(lft, l, m, x, y);
  }
  if (y > m && t[rgt].mx > 2) {
    update(rgt, m + 1, r, x, y);
  }
  t[i] = join(t[lft], t[rgt]);
}
long long query(int i, int l, int r, int x, int y) {
  if (x <= l && r <= y) {
    return t[i].sum;
  }
  int m = (l + r) >> 1;
  long long res = 0;
  if (x <= m) {
    res += query(i << 1, l, m, x, y);
  }
  if (y > m) {
    res += query(i << 1 | 1, m + 1, r, x, y);
  }
  return res;
}
int main() {
  for (long long i = 2; i < M; ++i) {
    if (sp[i]) {
      continue;
    }
    sp[i] = i;
    for (long long j = i * i; j < M; j += i) {
      if (!sp[j]) {
        sp[j] = i;
      }
    }
  }
  d[1] = 1;
  for (int i = 2; i < M; ++i) {
    d[i] = 1;
    int x = i;
    int lst = sp[x];
    int cnt = 0;
    while (x > 1) {
      if (lst != sp[x]) {
        d[i] *= cnt + 1;
        cnt = 1;
        lst = sp[x];
      } else {
        ++cnt;
      }
      x /= sp[x];
    }
    d[i] *= cnt + 1;
  }
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", arr + i);
  }
  build(1, 0, n - 1);
  while (m--) {
    int type, l, r;
    scanf("%d %d %d", &type, &l, &r);
    --l;
    --r;
    --type;
    if (!type) {
      update(1, 0, n - 1, l, r);
    } else {
      printf("%lld\n", query(1, 0, n - 1, l, r));
    }
  }
  return 0;
}