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 = 1e5 + 5;
const int INF = 1e9 + 7;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
int d[N * 10], a[N * 3];
int n, m;
long long ans;
struct node {
  int l, r, mid;
  long long sum, lazy;
} t[N * 3 * 4];
void pushup(int rt) {
  t[rt].sum = t[rt << 1].sum + t[rt << 1 | 1].sum;
  t[rt].lazy = max(t[rt << 1].lazy, t[rt << 1 | 1].lazy);
}
void build(int l, int r, int rt) {
  int m = (l + r) / 2;
  t[rt].mid = m, t[rt].l = l, t[rt].r = r;
  t[rt].sum = 0, t[rt].lazy = 0;
  if (l == r) {
    t[rt].sum = t[rt].lazy = a[l];
    return;
  }
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void update(int l, int r, int rt) {
  if (t[rt].lazy < 3) return;
  if (t[rt].l == t[rt].r) {
    t[rt].sum = t[rt].lazy = d[t[rt].sum];
    return;
  }
  if (l <= t[rt].mid) update(l, r, rt << 1);
  if (r > t[rt].mid) update(l, r, rt << 1 | 1);
  pushup(rt);
}
void query(int l, int r, int rt) {
  if (l <= t[rt].l && t[rt].r <= r) {
    ans += t[rt].sum;
    return;
  }
  if (l <= t[rt].mid) query(l, r, rt << 1);
  if (r > t[rt].mid) query(l, r, rt << 1 | 1);
  pushup(rt);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, n, 1);
  for (int i = 1; i <= 1e6; i++) {
    int g = 1;
    while (g) {
      if (i * g > 1e6) break;
      d[i * g]++;
      g++;
    }
  }
  while (m--) {
    int t, l, r;
    scanf("%d%d%d", &t, &l, &r);
    if (t == 1) update(l, r, 1);
    if (t == 2) {
      ans = 0;
      query(l, r, 1);
      printf("%lld\n", ans);
    }
  }
}