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 arr[300005];
int Cache[1000005];
pair<long long, int> segTree[1200005];
int Divisors(int n) {
  int cnt = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return Cache[n] = cnt;
}
int buildTree(int pos, int l, int r) {
  if (l == r) {
    segTree[pos].first = arr[l];
    segTree[pos].second = (arr[l] <= 2);
    return 0;
  }
  int mid = (l + r) / 2;
  buildTree(pos * 2, l, mid);
  buildTree(pos * 2 + 1, mid + 1, r);
  segTree[pos].first = segTree[pos * 2].first + segTree[pos * 2 + 1].first;
  segTree[pos].second = segTree[pos * 2].second + segTree[pos * 2 + 1].second;
  return 0;
}
long long sum(int pos, int l, int r, int UL, int UR) {
  if (l > UR || r < UL) return 0;
  if (l >= UL && r <= UR) return segTree[pos].first;
  int mid = (l + r) / 2;
  return sum(pos * 2, l, mid, UL, UR) + sum(pos * 2 + 1, mid + 1, r, UL, UR);
}
void rep(int pos, int l, int r, int UL, int UR) {
  if (l > UR || r < UL) return;
  if (l >= UL && r <= UR) {
    if (r - l + 1 == segTree[pos].second) return;
    if (l == r) {
      segTree[pos].first = Divisors(segTree[pos].first);
      segTree[pos].second = (segTree[pos].first <= 2);
      return;
    }
  }
  int mid = (l + r) / 2;
  rep(pos * 2, l, mid, UL, UR);
  rep(pos * 2 + 1, mid + 1, r, UL, UR);
  segTree[pos].first = segTree[pos * 2].first + segTree[pos * 2 + 1].first;
  segTree[pos].second = segTree[pos * 2].second + segTree[pos * 2 + 1].second;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  buildTree(1, 1, n);
  while (m--) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    if (t == 1)
      rep(1, 1, n, l, r);
    else
      printf("%lld\n", sum(1, 1, n, l, r));
  }
}