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 maxn = 1e6 + 1;
int d[maxn];
void sieve() {
  for (int i = 1; i < maxn; ++i) {
    for (int j = i; j < maxn; j += i) d[j]++;
  }
}
long long t_sum[1200001];
bool flag[1200001];
int a[maxn];
inline void merge(int v) {
  t_sum[v] = t_sum[v << 1] + t_sum[v << 1 | 1];
  flag[v] = flag[v << 1] & flag[v << 1 | 1];
}
inline bool p(int a) { return a <= 2; }
int tl[1200001];
int tr[1200001];
void build(int v, int _tl, int _tr) {
  tl[v] = _tl;
  tr[v] = _tr;
  if (_tl == _tr - 1) {
    t_sum[v] = a[_tl];
    flag[v] = p(a[_tl]);
    return;
  }
  int tm = (_tl + _tr) >> 1;
  build(v << 1, _tl, tm);
  build(v << 1 | 1, tm, _tr);
  merge(v);
}
int L, R;
long long get_sum(int v) {
  if (tl[v] >= R || L >= tr[v]) return 0;
  if (tl[v] >= L && tr[v] <= R) return t_sum[v];
  return get_sum(v << 1) + get_sum(v << 1 | 1);
}
void upd(int v) {
  if (tl[v] >= R || L >= tr[v]) {
    return;
  }
  if (tl[v] == tr[v] - 1) {
    a[tl[v]] = d[a[tl[v]]];
    t_sum[v] = a[tl[v]];
    flag[v] = p(a[tl[v]]);
    return;
  }
  if (tl[v] >= L && tr[v] <= R && flag[v]) {
    return;
  }
  upd(v << 1);
  upd(v << 1 | 1);
  merge(v);
}
int main() {
  sieve();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  build(1, 1, n + 1);
  while (m--) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int l, r;
      scanf("%d%d", &l, &r);
      L = l;
      R = r + 1;
      upd(1);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      L = l;
      R = r + 1;
      printf("%I64d\n", get_sum(1));
    }
  }
}