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;
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * w;
}
int d[1000010], t[1000010], prime[1000010], tot, vis[1000010];
int n, fa[1000010], a[1000010], m;
long long c[1000010];
inline void prepare(int Maxn) {
  d[1] = 1;
  for (register int i = 2; i <= Maxn; i++) {
    if (!vis[i]) {
      prime[++tot] = i;
      t[i] = d[i] = 2;
    }
    for (register int j = 1; 1LL * prime[j] * i <= Maxn; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j]) {
        d[i * prime[j]] = d[i] * 2;
        t[i * prime[j]] = 2;
      } else {
        d[i * prime[j]] = d[i] / t[i] * (t[i] + 1);
        t[i * prime[j]] = t[i] + 1;
        break;
      }
    }
  }
}
inline void add(int x, int v) {
  for (; x <= n; x += x & (-x)) c[x] += v;
}
inline long long query(int x) {
  long long sum = 0;
  for (; x; x -= x & (-x)) sum += c[x];
  return sum;
}
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  prepare(1000010 - 10);
  n = read(), m = read();
  for (register int i = 1; i <= n; i++) a[i] = read(), fa[i] = i;
  fa[n + 1] = n + 1;
  for (register int i = 1; i <= n; i++) add(i, a[i]);
  while (m--) {
    int opt = read(), l = read(), r = read();
    if (opt == 1) {
      for (register int i = l; i <= r; i = find(i + 1)) {
        add(i, d[a[i]] - a[i]);
        a[i] = d[a[i]];
        if (a[i] == 2 || a[i] == 1) fa[find(i)] = find(i + 1);
      }
    } else {
      printf("%lld\n", query(r) - query(l - 1));
    }
  }
  return 0;
}