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;
using ll = long long;
const int N = 3E5 + 7;
ll sum[N << 2], sum1[N << 2];
int mp[1000009];
void push(int x) {
  sum[x] = sum[x << 1] + sum[x << 1 | 1];
  sum1[x] = sum1[x << 1] + sum1[x << 1 | 1];
}
void build(int l, int r, int rt) {
  if (l == r) {
    cin >> sum[rt];
    if (sum[rt] == 1) sum1[rt]++;
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  push(rt);
}
void update(int L, int R, int l, int r, int rt) {
  if (sum[rt] + sum1[rt] == (r - l + 1ll) * 2) return;
  if (l == r) {
    sum[rt] = mp[sum[rt]];
    return;
  }
  int mid = (l + r) >> 1;
  if (L <= mid) {
    update(L, R, l, mid, rt << 1);
  }
  if (R > mid) {
    update(L, R, mid + 1, r, rt << 1 | 1);
  }
  push(rt);
}
ll query(int L, int R, int l, int r, int rt) {
  if (L <= l && R >= r) {
    return sum[rt];
  }
  ll ans = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ans += query(L, R, l, mid, rt << 1);
  if (R > mid) ans += query(L, R, mid + 1, r, rt << 1 | 1);
  return ans;
}
void linit() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) {
      mp[j]++;
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  linit();
  int n, m;
  cin >> n >> m;
  build(1, n, 1);
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    if (x == 1) {
      update(y, z, 1, n, 1);
    } else {
      cout << query(y, z, 1, n, 1) << endl;
    }
  }
}
