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 V = 1e6 + 10, N = 3e5 + 10;
int no[V];
bool same[4 * N];
long long sum[4 * N];
int arr[N];
void build(int idx, int s, int e) {
  if (s == e) {
    if (no[arr[s]] == arr[s]) same[idx] = 1;
    sum[idx] = arr[s];
    return;
  }
  int m = s + ((e - s) >> 1), L = (idx << 1) | 1, R = L + 1;
  build(L, s, m);
  build(R, m + 1, e);
  sum[idx] = sum[L] + sum[R];
  same[idx] = (same[L] && same[R]);
}
void replace(int idx, int s, int e, int l, int r) {
  if (e < l || s > r || same[idx]) return;
  if (s == e) {
    sum[idx] = no[sum[idx]];
    if (no[sum[idx]] == sum[idx]) same[idx] = 1;
    return;
  }
  int m = s + ((e - s) >> 1), L = (idx << 1) | 1, R = L + 1;
  replace(L, s, m, l, r);
  replace(R, m + 1, e, l, r);
  sum[idx] = sum[L] + sum[R];
  same[idx] = (same[L] && same[R]);
}
long long getSum(int idx, int s, int e, int l, int r) {
  if (e < l || s > r) return 0;
  if (s >= l && e <= r) return sum[idx];
  int m = s + ((e - s) >> 1), L = (idx << 1) | 1, R = L + 1;
  return getSum(L, s, m, l, r) + getSum(R, m + 1, e, l, r);
}
int n, m;
int main() {
  for (int i = 1; i < V; i++)
    for (int j = i; j < V; j += i) ++no[j];
  scanf("%d", &n), scanf("%d", &m);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]);
  build(0, 0, n - 1);
  for (int i = 0; i < m; ++i) {
    int t, l, r;
    scanf("%d", &t), scanf("%d", &l), scanf("%d", &r);
    l--, r--;
    if (t == 1) {
      replace(0, 0, n - 1, l, r);
    } else {
      printf("%lld\n", getSum(0, 0, n - 1, l, r));
    }
  }
}