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 = 1e6 + 5;
const int MAXN = 3e5 + 100;
int cnt[maxn];
int ans[MAXN << 2], a[MAXN];
long long s[MAXN << 2];
int n, q;
void build(int idx, int x, int y) {
  if (x == y) {
    s[idx] = ans[idx] = a[x];
    return;
  }
  int mid = (x + y) >> 1;
  build(idx << 1, x, mid);
  build(idx << 1 | 1, mid + 1, y);
  ans[idx] = max(ans[idx << 1], ans[idx << 1 | 1]);
  s[idx] = s[idx << 1] + s[idx << 1 | 1];
}
void replace(int l, int r, int idx, int x, int y) {
  if (l > y or x > r or ans[idx] <= 2) return;
  if (x == y) {
    a[x] = cnt[a[x]];
    s[idx] = ans[idx] = a[x];
    return;
  }
  int mid = (x + y) >> 1;
  replace(l, r, idx << 1, x, mid);
  replace(l, r, idx << 1 | 1, mid + 1, y);
  ans[idx] = max(ans[idx << 1], ans[idx << 1 | 1]);
  s[idx] = s[idx << 1] + s[idx << 1 | 1];
}
long long cal(int l, int r, int idx, int x, int y) {
  if (l > y or x > r) return 0;
  if (l <= x and y <= r) return s[idx];
  int mid = (x + y) >> 1;
  return cal(l, r, idx << 1, x, mid) + cal(l, r, idx << 1 | 1, mid + 1, y);
}
void solve() {
  build(1, 1, n);
  while (q--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1)
      replace(l, r, 1, 1, n);
    else
      printf("%I64d\n", cal(l, r, 1, 1, n));
  }
}
void input() {
  for (int i = 1; i < maxn; i++) {
    for (int j = i; j < maxn; j += i) cnt[j]++;
  }
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
}
int main() {
  input();
  solve();
  return 0;
}