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>
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
constexpr long long MAX = 5100000;
constexpr long long INF = 1LL << 60;
constexpr int inf = 1 << 28;
constexpr long long mod = 1000000007LL;
using namespace std;
class BIT {
 private:
  int N;
  vector<long long> node;

 public:
  BIT(int M) {
    node = vector<long long>(M + 1, 0);
    N = M;
  }
  long long sum1(int i) {
    long long s = 0;
    while (i > 0) {
      s += node[i];
      i -= i & -i;
    }
    return s;
  }
  long long sum1(int i, int j) {
    long long l = sum1(i - 1);
    long long r = sum1(j);
    return r - l;
  }
  void add1(int i, long long x) {
    while (i <= N) {
      node[i] += x;
      i += i & -i;
    }
  }
  long long sum(int left, int right) { return sum1(left + 1, right); }
  void add(int i, long long x) { add1(i + 1, x); }
  void update(int i, long long x) {
    long long s = sum(i, i + 1);
    add(i, x - s);
  }
};
long long cnv[1000001];
int main() {
  long long n, m;
  scanf("%lld %lld", &n, &m);
  vector<long long> a(n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  BIT bit(n);
  for (int i = 0; i < n; i++) bit.add(i, a[i]);
  for (int i = 1; i <= 1000000; i++)
    for (int j = i; j <= 1000000; j += i) cnv[j]++;
  set<int> s;
  s.emplace(n);
  for (int i = 0; i < n; i++) {
    if (a[i] > 2) s.emplace(i);
  }
  while (m--) {
    int t, l, r;
    scanf("%d %d %d", &t, &l, &r);
    l--;
    if (t == 1) {
      for (int i = l; i < r;) {
        a[i] = cnv[a[i]];
        if (a[i] <= 2 and s.find(i) != s.end()) s.erase(s.find(i));
        bit.update(i, a[i]);
        i = *s.upper_bound(i);
      }
    } else {
      printf("%lld\n", bit.sum(l, r));
    }
  }
  return 0;
}