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;
int num[1000100], cnt, a[300010];
void init() {
  memset(num, 0, sizeof(num));
  for (int i = 1; i < 1000100; i++) {
    for (int j = i; j < 1000100; j += i) {
      num[j]++;
    }
  }
}
struct tree {
  int l, r, mx;
  long long sum;
} tr[4 * 300010];
void push(int d) {
  tr[d].mx = max(tr[d * 2].mx, tr[d * 2 + 1].mx);
  tr[d].sum = tr[d * 2].sum + tr[d * 2 + 1].sum;
}
void build(int d, int l, int r) {
  tr[d].l = l, tr[d].r = r;
  if (l == r) {
    tr[d].sum = tr[d].mx = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(2 * d, l, mid);
  build(2 * d + 1, mid + 1, r);
  push(d);
}
void update(int d, int L, int R) {
  if (tr[d].l >= L && tr[d].r <= R) {
    if (tr[d].mx <= 2)
      return;
    else if (tr[d].l == tr[d].r) {
      tr[d].mx = tr[d].sum = num[tr[d].mx];
      return;
    }
  }
  int mid = (tr[d].l + tr[d].r) / 2;
  if (R <= mid)
    update(d * 2, L, R);
  else if (L > mid)
    update(d * 2 + 1, L, R);
  else
    update(d * 2, L, mid), update(d * 2 + 1, mid + 1, R);
  push(d);
}
long long query(int d, int L, int R) {
  if (tr[d].l == L && tr[d].r == R) {
    return tr[d].sum;
  }
  int mid = (tr[d].l + tr[d].r) / 2;
  if (R <= mid)
    return query(d * 2, L, R);
  else if (L > mid)
    return query(d * 2 + 1, L, R);
  else
    return query(d * 2, L, mid) + query(d * 2 + 1, mid + 1, R);
}
int main() {
  init();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  while (m--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) {
      update(1, l, r);
    } else {
      printf("%lld\n", query(1, l, r));
    }
  }
}