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 = 1e6 + 5;
int a[N], d[N], n, m, fa[N];
struct Tree {
  int l, r, flag;
  long long val;
} tr[N * 4];
void pushUp(int t) {
  tr[t].val = tr[t << 1].val + tr[(t << 1) | 1].val;
  tr[t].flag = tr[t << 1].flag && tr[(t << 1) | 1].flag;
}
void build(int t, int l, int r) {
  tr[t].l = l;
  tr[t].r = r;
  if (l == r) {
    tr[t].val = a[l];
    fa[l] = t;
    return;
  }
  int mid = (l + r) / 2;
  build(t << 1, l, mid);
  build((t << 1) | 1, mid + 1, r);
  pushUp(t);
}
void update(int t, int l, int r) {
  if (tr[t].l == tr[t].r) {
    tr[t].val = d[tr[t].val];
    if (tr[t].val <= 2) tr[t].flag = 1;
    return;
  }
  if (tr[t].flag == 1) return;
  int mid = (tr[t].l + tr[t].r) / 2;
  if (mid >= l) update(t << 1, l, r);
  if (mid < r) update((t << 1) | 1, l, r);
  pushUp(t);
}
long long query(int t, int l, int r) {
  if (tr[t].l >= l && tr[t].r <= r) {
    return tr[t].val;
  }
  long long ret = 0;
  int mid = (tr[t].l + tr[t].r) / 2;
  if (mid >= l) ret += query(t << 1, l, r);
  if (mid < r) ret += query((t << 1) | 1, l, r);
  return ret;
}
void solve() {
  for (int i = 1; i < N; i++) {
    for (int j = i; j < N; j += i) d[j]++;
  }
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  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, l, r);
    if (op == 2) printf("%lld\n", query(1, l, r));
  }
}
int main() {
  solve();
  return 0;
}