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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
const int N = 300005, M = 1000005;
int n, m, cnt[M], a[N];
void sieve(int l) {
  for (int i = 1; i <= l; i++)
    for (int j = i; j <= l; j += i) cnt[j]++;
}
struct SegmentTree {
  long long t[N << 2];
  int mx[N << 2];
  void build(int rt, int l, int r) {
    if (l == r) {
      t[rt] = mx[rt] = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    t[rt] = t[rt << 1] + t[rt << 1 | 1],
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
  }
  void modify(int rt, int l, int r, int ql, int qr) {
    if (mx[rt] <= 2) return;
    if (l == r) {
      a[l] = cnt[a[l]];
      t[rt] = mx[rt] = a[l];
      return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) modify(rt << 1, l, mid, ql, qr);
    if (qr > mid) modify(rt << 1 | 1, mid + 1, r, ql, qr);
    t[rt] = t[rt << 1] + t[rt << 1 | 1],
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
  }
  long long ask(int rt, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr) return t[rt];
    int mid = (l + r) >> 1;
    long long ret = 0;
    if (ql <= mid) ret += ask(rt << 1, l, mid, ql, qr);
    if (qr > mid) ret += ask(rt << 1 | 1, mid + 1, r, ql, qr);
    return ret;
  }
} st;
int main() {
  sieve(1e6);
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  st.build(1, 1, n);
  while (m--) {
    int opt = read(), l = read(), r = read();
    if (opt == 1)
      st.modify(1, 1, n, l, r);
    else
      printf("%lld\n", st.ask(1, 1, n, l, r));
  }
  return 0;
}
