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 N = 3e5 + 5;
const int M = 1e6 + 5;
long long tree[N << 2];
bool ok[N << 2];
int num[M], a[N];
int n, m;
void PushUp(int rt) {
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
  ok[rt] = ok[rt << 1] && ok[rt << 1 | 1];
}
void Build(int l, int r, int rt) {
  if (l == r) {
    tree[rt] = a[l];
    if (tree[rt] == 1 || tree[rt] == 2)
      ok[rt] = 1;
    else
      ok[rt] = 0;
    return;
  }
  int m = l + r >> 1;
  Build(l, m, rt << 1);
  Build(m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
long long Query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return tree[rt];
  int m = l + r >> 1;
  long long ans = 0;
  if (L <= m) ans += Query(L, R, l, m, rt << 1);
  if (m < R) ans += Query(L, R, m + 1, r, rt << 1 | 1);
  return ans;
}
void Revise(int L, int R, int l, int r, int rt) {
  if (ok[rt]) return;
  if (l == r) {
    tree[rt] = num[tree[rt]];
    if (tree[rt] == 1 || tree[rt] == 2) ok[rt] = 1;
    return;
  }
  int m = l + r >> 1;
  if (L <= m) Revise(L, R, l, m, rt << 1);
  if (m < R) Revise(L, R, m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(num, 0, sizeof(num));
  for (int i = 1; i < M; i++)
    for (int j = i; j < M; j += i) num[j]++;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  Build(1, n, 1);
  int q, l, r;
  while (m--) {
    cin >> q >> l >> r;
    if (q == 2)
      cout << Query(l, r, 1, n, 1) << endl;
    else
      Revise(l, r, 1, n, 1);
  }
  return 0;
}