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 a[(1 << 21)], n, q;
int it[(1 << 21)];
long long sum[(1 << 21)], k = 19;
int d[(1 << 21)];
int cmp(int ai, int b) { return a[ai] > a[b] ? ai : b; }
void upd(int i, int v) {
  a[i] = v;
  i += (1 << k);
  it[i] = i - (1 << k);
  sum[i] = v;
  for (i /= 2; i; i /= 2) {
    sum[i] = sum[i * 2] + sum[i * 2 + 1];
    it[i] = cmp(it[i * 2], it[i * 2 + 1]);
  }
}
int getMaxInd(int l, int r, int ll = 0, int rr = (1 << k), int ind = 1) {
  if (l <= ll && rr <= r) {
    return it[ind];
  }
  if (rr <= l || r <= ll) {
    return n;
  }
  return cmp(getMaxInd(l, r, ll, (ll + rr) / 2, ind * 2),
             getMaxInd(l, r, (ll + rr) / 2, rr, ind * 2 + 1));
}
long long getSum(int l, int r, int ll = 0, int rr = (1 << k), int ind = 1) {
  if (l <= ll && rr <= r) {
    return sum[ind];
  }
  if (rr <= l || r <= ll) {
    return 0;
  }
  return getSum(l, r, ll, (ll + rr) / 2, ind * 2) +
         getSum(l, r, (ll + rr) / 2, rr, ind * 2 + 1);
}
stack<pair<int, long long> > toUpd;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> q;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    upd(i, a[i]);
  }
  for (long long i = 1; i <= a[it[1]]; i++) {
    for (long long j = 1; i * j <= a[it[1]]; j++) {
      d[i * j]++;
    }
  }
  for (long long i = 0; i < q; i++) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long l, r;
      cin >> l >> r;
      l--;
      for (;;) {
        long long cInd = getMaxInd(l, r);
        if (d[a[cInd]] == a[cInd]) {
          break;
        }
        toUpd.push(make_pair(cInd, a[cInd]));
        upd(cInd, 0);
      }
      for (; !toUpd.empty();) {
        long long cInd = toUpd.top().first;
        long long val = toUpd.top().second;
        toUpd.pop();
        upd(cInd, d[val]);
      }
    } else {
      long long l, r;
      cin >> l >> r;
      l--;
      cout << getSum(l, r) << "\n";
    }
  }
}
