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 long long maxn = 3e5 + 10;
long long n, m, a[maxn], d[1000010];
struct seg_tree {
  struct nodes {
    long long l, r, sum, max_num;
  } node[maxn << 2];
  void up(long long now) {
    return (
        void)(node[now].sum = node[(now << 1)].sum + node[(now << 1 | 1)].sum,
              node[now].max_num =
                  max(node[(now << 1)].max_num, node[(now << 1 | 1)].max_num));
  }
  void build(long long l, long long r, long long now) {
    node[now].l = l, node[now].r = r;
    if (l == r) return (void)(node[now].sum = a[l], node[now].max_num = a[l]);
    build(l, ((l + r) >> 1), (now << 1)),
        build(((l + r) >> 1) + 1, r, (now << 1 | 1));
    return up(now);
  }
  void chenge(long long l, long long r, long long now) {
    if (r < node[now].l or node[now].r < l or node[now].max_num <= 2) return;
    if (node[now].l == node[now].r)
      return (void)(node[now].max_num = node[now].sum = d[node[now].sum]);
    chenge(l, r, (now << 1)), chenge(l, r, (now << 1 | 1));
    return up(now);
  }
  void quary(long long l, long long r, long long now, long long &ans) {
    if (r < node[now].l or node[now].r < l) return;
    if (l <= node[now].l and node[now].r <= r)
      return (void)(ans += node[now].sum);
    quary(l, r, (now << 1), ans), quary(l, r, (now << 1 | 1), ans);
    return up(now);
  }
} tree;
signed main() {
  for (long long i = 1; i <= 1000000; i++) {
    for (long long j = i; j <= 1000000; j += i) {
      d[j]++;
    }
  }
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  tree.build(1, n, 1);
  for (long long i = 1, x, y, z; i <= m; i++) {
    scanf("%lld%lld%lld", &x, &y, &z);
    if (x & 1)
      tree.chenge(y, z, 1);
    else {
      long long ans = 0;
      tree.quary(y, z, 1, ans);
      printf("%lld\n", ans);
    }
  }
  return 0;
}