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>
const int M = 3e5 + 10;
const int N = 1e6 + 10;
int num[N];
long long T[M << 2], done[M << 2];
void up(int o) {
  T[o] = T[o << 1] + T[o << 1 | 1];
  done[o] = std::max(done[o << 1], done[o << 1 | 1]);
}
void build(int o, int l, int r) {
  if (l == r) {
    scanf("%lld", &T[o]);
    done[o] = T[o];
    return;
  }
  int mid = (l + r) >> 1;
  build(o << 1, l, mid);
  build(o << 1 | 1, mid + 1, r);
  up(o);
}
void update(int o, int l, int r, int ql, int qr) {
  if (done[o] <= 2) return;
  if (l == r) {
    T[o] = done[o] = num[T[o]];
    return;
  }
  int mid = (l + r) >> 1;
  if (ql <= mid) update(o << 1, l, mid, ql, qr);
  if (qr > mid) update(o << 1 | 1, mid + 1, r, ql, qr);
  up(o);
}
long long ask(int o, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return T[o];
  long long ret = 0;
  int mid = (l + r) >> 1;
  if (ql <= mid) ret += ask(o << 1, l, mid, ql, qr);
  if (qr > mid) ret += ask(o << 1 | 1, mid + 1, r, ql, qr);
  return ret;
}
int main() {
  for (int i = 1; i < N; ++i) {
    for (int j = i; j < N; j += i) {
      num[j]++;
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  for (int i = 0; i < m; ++i) {
    int op, l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1)
      update(1, 1, n, l, r);
    else
      printf("%lld\n", ask(1, 1, n, l, r));
  }
}
