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;
long long num[1000001], d[1000001], n, m, pri[1000001], cnt, a[300001],
    t[300001], fa[300001];
bool flag[1000001];
inline void pre() {
  flag[1] = true;
  d[1] = 1;
  for (int i = 2; i <= 1000000; i++) {
    if (!flag[i]) {
      pri[++cnt] = i;
      d[i] = 2;
      num[i] = 1;
    }
    for (int j = 1; j <= cnt && i * pri[j] <= 1000000; j++) {
      flag[i * pri[j]] = true;
      if (i % pri[j]) {
        d[i * pri[j]] = d[i] * 2;
        num[i * pri[j]] = 1;
      } else {
        d[i * pri[j]] = d[i] / (num[i] + 1) * (num[i] + 2);
        num[i * pri[j]] = num[i] + 1;
        break;
      }
    }
  }
}
inline long long find(long long x) {
  if (x == fa[x])
    return x;
  else
    return fa[x] = find(fa[x]);
}
inline long long lowbit(long long x) { return x & (-x); }
inline void update(long long x, long long k) {
  while (x <= n) {
    t[x] += k;
    x += lowbit(x);
  }
}
inline long long query(long long x) {
  long long res = 0;
  while (x) {
    res += t[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  scanf("%lld%lld", &n, &m);
  pre();
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    update(i, a[i]);
    fa[i] = i;
  }
  fa[n + 1] = n + 1;
  for (int i = 1; i <= m; i++) {
    long long op, l, r;
    scanf("%lld%lld%lld", &op, &l, &r);
    if (l > r) swap(l, r);
    if (op == 1)
      while (l <= r) {
        long long t = d[a[l]];
        update(l, t - a[l]);
        a[l] = t;
        if (a[l] <= 2) fa[l] = l + 1;
        if (fa[l] == l)
          l++;
        else
          l = find(l);
      }
    if (op == 2) printf("%lld\n", query(r) - query(l - 1));
  }
  return 0;
}