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 MAXN = 524288;
const int MAXA = 1000001;
int d[MAXA + 1];
int cnt[MAXA + 1];
int segmax[2 * MAXN + 1];
long long segsum[2 * MAXN + 1];
int segl[2 * MAXN + 1];
int segr[2 * MAXN + 1];
void initialize() {
  for (int i = 1; i <= MAXA; i++) {
    int j = i;
    while (j <= MAXA) {
      d[j]++;
      j += i;
    }
  }
  for (int i = 3; i <= MAXA; i++) {
    cnt[i] = cnt[d[i]] + 1;
  }
}
void build() {
  for (int i = MAXN; i < MAXN + MAXN; i++) {
    segl[i] = segr[i] = i - MAXN;
  }
  for (int i = MAXN - 1; i > 0; i--) {
    segsum[i] = segsum[i + i] + segsum[i + i + 1];
    segmax[i] = max(segmax[i + i], segmax[i + i + 1]);
    segl[i] = segl[2 * i];
    segr[i] = segr[2 * i + 1];
  }
}
void replace(int l, int r, int now) {
  if (segmax[now] == 0) {
    return;
  }
  if (now >= MAXN) {
    segsum[now] = d[segsum[now]];
    segmax[now]--;
    int t = now / 2;
    while (t > 0) {
      segsum[t] = segsum[t + t] + segsum[t + t + 1];
      segmax[t] = max(segmax[t + t], segmax[t + t + 1]);
      t /= 2;
    }
    return;
  }
  if (r <= segr[now + now]) {
    replace(l, r, now + now);
  } else if (l >= segl[now + now + 1]) {
    replace(l, r, now + now + 1);
  } else {
    replace(l, segr[now + now], now + now);
    replace(segl[now + now + 1], r, now + now + 1);
  }
}
void replace(int l, int r) { replace(l, r, 1); }
long long sum(int l, int r, int now) {
  if (l == segl[now] && r == segr[now]) {
    return segsum[now];
  }
  if (r <= segr[now + now]) {
    return sum(l, r, now + now);
  }
  if (l >= segl[now + now + 1]) {
    return sum(l, r, now + now + 1);
  }
  return sum(l, segr[now + now], now + now) +
         sum(segl[now + now + 1], r, now + now + 1);
}
long long sum(int l, int r) { return sum(l, r, 1); }
int main() {
  initialize();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &segsum[MAXN + i]);
    segmax[MAXN + i] = cnt[segsum[MAXN + i]];
  }
  build();
  for (int i = 0; i < m; i++) {
    int op;
    int l, r;
    scanf("%d%d%d", &op, &l, &r);
    if (op == 1) replace(l, r);
    if (op == 2) cout << sum(l, r) << endl;
  }
  return 0;
}