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>
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int maxm = 1e6 + 5;
int ttt[maxm];
const int maxn = 3e5 + 5;
int ori[maxn];
class t {
 public:
  int l, r, cnt, num, mode, len;
  long long sum;
  t() {}
} tree[maxn * 4 + 5];
void build(int l, int r, int u) {
  tree[u].l = l;
  tree[u].r = r;
  tree[u].len = r - l + 1;
  tree[u].mode = 1;
  if (l == r) {
    tree[u].sum = ori[l];
    if (tree[u].sum <= 2) tree[u].mode = 0;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, ((u << 1)));
  build(mid + 1, r, ((u << 1) | 1));
  tree[u].sum = tree[((u << 1))].sum + tree[((u << 1) | 1)].sum;
}
void update(int l, int r, int u) {
  if (tree[u].mode == 0) {
    return;
  }
  if (tree[u].len == 1) {
    tree[u].sum = ttt[tree[u].sum];
    if (tree[u].sum <= 2) tree[u].mode = 0;
    return;
  }
  int mid = (tree[u].r + tree[u].l) >> 1;
  if (l <= mid) update(l, r, ((u << 1)));
  if (r > mid) update(l, r, ((u << 1) | 1));
  tree[u].sum = tree[((u << 1))].sum + tree[((u << 1) | 1)].sum;
  tree[u].mode = tree[((u << 1))].mode | tree[((u << 1) | 1)].mode;
}
long long query(int l, int r, int u) {
  if (tree[u].l >= l && tree[u].r <= r) {
    return tree[u].sum;
  }
  int mid = (tree[u].r + tree[u].l) >> 1;
  long long ans = 0;
  if (l <= mid) ans += query(l, r, ((u << 1)));
  if (r > mid) ans += query(l, r, ((u << 1) | 1));
  return ans;
}
template <class T>
inline void scan_d(T &ret) {
  char c;
  ret = 0;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') {
    ret = ret * 10 + (c - '0'), c = getchar();
  }
}
int main() {
  int i, j, k, m, n, T;
  for (int i = 1; i <= 1000000; ++i)
    for (int j = i; j <= 1000000; j += i) ttt[j]++;
  scan_d(n);
  scan_d(m);
  for (int i = 1; i <= n; ++i) scan_d(ori[i]);
  build(1, n, 1);
  int ta, tb, tc;
  for (int i = 1; i <= m; ++i) {
    scan_d(ta);
    scan_d(tb);
    scan_d(tc);
    if (ta == 1)
      update(tb, tc, 1);
    else {
      long long ans = query(tb, tc, 1);
      printf("%lld\n", ans);
    }
  }
  return 0;
}
