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;
const int M = 1e6;
const long long LIM = 1e18;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const double PI = atan(1.0) * 4;
const int mod = 1e9 + 7;
int n, m, a[N + 10], d[M + 10];
int tree[(N + 10) << 2];
long long sum[(N + 10) << 2];
void pushup(int x) {
  tree[x] = max(tree[x << 1], tree[x << 1 | 1]);
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
}
void build(int x, int l, int r) {
  if (l == r) {
    tree[x] = sum[x] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(x << 1, l, mid);
  build(x << 1 | 1, mid + 1, r);
  pushup(x);
}
void update(int x, int l, int r, int L, int R) {
  if (tree[x] <= 2) return;
  if (l == r) {
    tree[x] = sum[x] = d[tree[x]];
    return;
  }
  int mid = (l + r) / 2;
  if (L <= mid) update(x << 1, l, mid, L, R);
  if (R > mid) update(x << 1 | 1, mid + 1, r, L, R);
  pushup(x);
}
long long query(int x, int l, int r, int L, int R) {
  if (L <= l && r <= R) return sum[x];
  int mid = (l + r) / 2;
  long long res = 0;
  if (L <= mid) res += query(x << 1, l, mid, L, R);
  if (R > mid) res += query(x << 1 | 1, mid + 1, r, L, R);
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= M; i++)
    for (int j = i; j <= M; j += i) d[j]++;
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int opt, l, r;
    cin >> opt >> l >> r;
    if (opt == 1)
      update(1, 1, n, l, r);
    else
      cout << query(1, 1, n, l, r) << "\n";
  }
  return 0;
}