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
#include <bits/stdc++.h>
using namespace std;
int n, m;
int d[1000005], ok[4 * 300005];
long long sum[4 * 300005];
void init() {
  memset(sum, 0, sizeof sum);
  memset(ok, 0, sizeof ok);
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) d[j]++;
  }
}
void build(int v, int l, int r) {
  if (l == r) {
    long long x;
    scanf("%lld", &sum[v]);
    if (sum[v] == 1 || sum[v] == 2) ok[v] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(v * 2, l, mid);
  build(v * 2 + 1, mid + 1, r);
  sum[v] = sum[v * 2] + sum[v * 2 + 1];
  ok[v] = ok[v * 2] & ok[v * 2 + 1];
}
void updata(int v, int l, int r, int ql, int qr) {
  if (ok[v]) return;
  if (l == r) {
    sum[v] = d[sum[v]];
    if (sum[v] == 1 || sum[v] == 2) ok[v] = 1;
    return;
  }
  int mid = (l + r) / 2;
  if (ql <= mid) updata(v * 2, l, mid, ql, qr);
  if (qr > mid) updata(v * 2 + 1, mid + 1, r, ql, qr);
  sum[v] = sum[v * 2] + sum[v * 2 + 1];
  ok[v] = ok[v * 2] & ok[v * 2 + 1];
}
long long query(int v, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[v];
  int mid = (l + r) / 2;
  long long s = 0;
  if (ql <= mid) s += query(v * 2, l, mid, ql, qr);
  if (qr > mid) s += query(v * 2 + 1, mid + 1, r, ql, qr);
  return s;
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  build(1, 1, n);
  int op, l, r;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1)
      updata(1, 1, n, l, r);
    else
      printf("%lld\n", query(1, 1, n, l, r));
  }
  return 0;
}
