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;
const int maxv = 1e6 + 5;
const int maxn = 3e5 + 5;
int cnt[maxv], node[maxn], mx[maxn << 2];
long long sum[maxn << 2];
void push_up(int n) {
  mx[n] = max(mx[n << 1], mx[n << 1 | 1]);
  sum[n] = sum[n << 1] + sum[n << 1 | 1];
}
void build(int l, int r, int n) {
  if (l == r) {
    mx[n] = sum[n] = node[l];
    return;
  }
  int m = (l + r) / 2;
  build(l, m, n << 1);
  build(m + 1, r, n << 1 | 1);
  push_up(n);
}
void update(int l, int r, int ql, int qr, int n) {
  if (mx[n] <= 2) return;
  if (l == r) {
    mx[n] = sum[n] = cnt[mx[n]];
    return;
  }
  int m = (l + r) / 2;
  if (qr <= m)
    update(l, m, ql, qr, n << 1);
  else if (ql > m)
    update(m + 1, r, ql, qr, n << 1 | 1);
  else {
    update(l, m, ql, m, n << 1);
    update(m + 1, r, m + 1, qr, n << 1 | 1);
  }
  push_up(n);
}
long long query(int l, int r, int ql, int qr, int n) {
  if (ql == l && qr == r) return sum[n];
  int m = (l + r) / 2;
  if (qr <= m) return query(l, m, ql, qr, n << 1);
  if (ql > m) return query(m + 1, r, ql, qr, n << 1 | 1);
  return query(l, m, ql, m, n << 1) + query(m + 1, r, m + 1, qr, n << 1 | 1);
}
int main() {
  for (int i = 1; i < maxv; i++) {
    for (int j = i; j < maxv; j += i) {
      cnt[j]++;
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &node[i]);
  build(1, n, 1);
  int q, l, r;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &q, &l, &r);
    if (q == 1)
      update(1, n, l, r, 1);
    else
      printf("%lld\n", query(1, n, l, r, 1));
  }
  return 0;
}
