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;
struct node {
  long long sum, max;
} a[maxn << 2];
int d[maxn];
void btree(int l, int r, int pos) {
  if (l == r) {
    scanf("%lld", &a[pos].sum);
    a[pos].max = a[pos].sum;
    return;
  }
  int mid = (l + r) >> 1;
  btree(l, mid, pos << 1);
  btree(mid + 1, r, pos << 1 | 1);
  a[pos].sum = a[pos << 1].sum + a[pos << 1 | 1].sum;
  a[pos].max = max(a[pos << 1].max, a[pos << 1 | 1].max);
}
void updata(int L, int R, int pos, int l, int r) {
  if (a[pos].max <= 2) return;
  if (L == R) {
    a[pos].sum = a[pos].max = d[a[pos].sum];
    return;
  }
  int mid = (L + R) >> 1;
  if (l <= mid) updata(L, mid, pos << 1, l, r);
  if (r > mid) updata(mid + 1, R, pos << 1 | 1, l, r);
  a[pos].sum = a[pos << 1].sum + a[pos << 1 | 1].sum;
  a[pos].max = max(a[pos << 1].max, a[pos << 1 | 1].max);
}
long long query(int L, int R, int pos, int l, int r) {
  if (l <= L && R <= r) {
    return a[pos].sum;
  }
  int mid = (L + R) >> 1;
  long long ans = 0;
  if (l <= mid) ans += query(L, mid, pos << 1, l, r);
  if (r > mid) ans += query(mid + 1, R, pos << 1 | 1, l, r);
  return ans;
}
int main() {
  int n, m, t, l, r;
  for (int i = 1; i <= maxn; i++)
    for (int j = i; j <= maxn; j += i) d[j]++;
  scanf("%d %d", &n, &m);
  btree(1, n, 1);
  while (m--) {
    scanf("%d %d %d", &t, &l, &r);
    if (t == 1)
      updata(1, n, 1, l, r);
    else
      printf("%lld\n", query(1, n, 1, l, r));
  }
}