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 n, m;
int f[1000010], vis[1000010], prime[1000010], d[1000010], a[300010], cnt;
void init() {
  f[1] = 1;
  for (int i = 2; i <= 1e6; i++) {
    if (!vis[i]) {
      prime[++cnt] = i;
      f[i] = 2;
      d[i] = 1;
    }
    for (int j = 1; j <= cnt && i * prime[j] <= 1e6; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        d[i * prime[j]] = d[i] + 1;
        f[i * prime[j]] = f[i] / (d[i] + 1) * (d[i * prime[j]] + 1);
        break;
      } else {
        d[i * prime[j]] = 1;
        f[i * prime[j]] = f[i] * f[prime[j]];
      }
    }
  }
}
struct tree {
  int l, r;
  long long sum, min1;
} tr[1200010];
void push_up(int root) {
  tr[root].sum = tr[root << 1].sum + tr[root << 1 | 1].sum;
  tr[root].min1 = max(tr[root << 1].min1, tr[root << 1 | 1].min1);
}
void build(int root, int l, int r) {
  if (l == r) {
    tr[root].l = l;
    tr[root].r = r;
    tr[root].sum = a[l];
    tr[root].min1 = a[l];
    return;
  }
  tr[root].l = l;
  tr[root].r = r;
  int mid = (l + r) >> 1;
  build(root << 1, l, mid);
  build(root << 1 | 1, mid + 1, r);
  push_up(root);
}
void update(int root, int l, int r) {
  if (tr[root].min1 <= 2) return;
  if (tr[root].l == tr[root].r) {
    tr[root].sum = f[tr[root].sum];
    tr[root].min1 = tr[root].sum;
    return;
  }
  int mid = (tr[root].l + tr[root].r) >> 1;
  if (r <= mid)
    update(root << 1, l, r);
  else if (mid < l)
    update(root << 1 | 1, l, r);
  else
    update(root << 1, l, mid), update(root << 1 | 1, mid + 1, r);
  push_up(root);
}
long long query(int root, int l, int r) {
  if (l <= tr[root].l && tr[root].r <= r) {
    return tr[root].sum;
  }
  int mid = (tr[root].l + tr[root].r) >> 1;
  if (r <= mid)
    return query(root << 1, l, r);
  else if (mid < l)
    return query(root << 1 | 1, l, r);
  else
    return query(root << 1, l, mid) + query(root << 1 | 1, mid + 1, r);
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int kd, l, r;
    scanf("%d%d%d", &kd, &l, &r);
    if (kd == 2) printf("%lld\n", query(1, l, r));
    if (kd == 1) update(1, l, r);
  }
}