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
#include <bits/stdc++.h>
using namespace std;
int cnt[1000010] = {};
void pre() {
  for (int i = 1; i < 1000010; i++) {
    for (int j = i; j < 1000010; j += i) {
      cnt[j]++;
    }
  }
}
long long int tree[2 * 1000010];
int n;
void build() {
  for (int i = n - 1; i > 0; i--) {
    tree[i] = tree[i << 1] + tree[(i << 1) | 1];
  }
}
void update(int idx, long long int val) {
  tree[idx + n] = val;
  for (int i = idx + n; i > 0; i >>= 1) {
    tree[i >> 1] = tree[i] + tree[i ^ 1];
  }
}
long long int query(int l, int r) {
  long long int ans = 0;
  l += n;
  r += n;
  for (; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      ans += tree[l];
      l++;
    }
    if (r & 1) {
      r--;
      ans += tree[r];
    }
  }
  return ans;
}
int main() {
  pre();
  int m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &tree[n + i]);
  }
  build();
  set<int> s;
  for (int i = 0; i < n; i++) {
    if (tree[n + i] > 2) {
      s.insert(i);
    }
  }
  while (m--) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l--;
    r--;
    if (t == 2) {
      printf("%lld\n", query(l, r + 1));
    } else {
      l--;
      vector<int> v;
      if (s.empty()) {
        continue;
      }
      if (s.upper_bound(l) == s.end()) {
        continue;
      }
      int idx = *s.upper_bound(l);
      while (idx <= r) {
        s.erase(idx);
        long long int x = tree[n + idx];
        update(idx, (long long int)cnt[x]);
        if (cnt[x] > 2) {
          v.push_back(idx);
        }
        if (s.empty()) {
          break;
        }
        if (s.upper_bound(l) == s.end()) {
          break;
        }
        idx = *s.upper_bound(l);
      }
      for (int u : v) {
        s.insert(u);
      }
    }
  }
  return 0;
}
