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 = 1000000 + 5;
long long tree[maxn * 4];
int a[maxn], skip[maxn * 4];
int d[1000001];
void init() {
  for (int i = 1; i <= 1000000; i++) {
    for (int j = i; j <= 1000000; j += i) d[j]++;
  }
}
void push_up(int rt) {
  tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
  skip[rt] = skip[rt << 1] && skip[rt << 1 | 1];
}
void build(int l, int r, int rt) {
  if (l == r) {
    tree[rt] = a[l];
    if (tree[rt] <= 2) skip[rt] = 1;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  push_up(rt);
}
long long query(int l, int r, int rt, int L, int R) {
  if (L <= l && r <= R) {
    return tree[rt];
  }
  int m = (l + r) >> 1;
  long long ans = 0;
  if (L <= m) ans += query(l, m, rt << 1, L, R);
  if (R > m) ans += query(m + 1, r, rt << 1 | 1, L, R);
  return ans;
}
void update(int l, int r, int rt, int L, int R) {
  if (skip[rt]) return;
  if (l == r) {
    tree[rt] = d[tree[rt]];
    if (tree[rt] <= 2) skip[rt] = 1;
    return;
  }
  int m = (l + r) >> 1;
  if (L <= m) update(l, m, rt << 1, L, R);
  if (R > m) update(m + 1, r, rt << 1 | 1, L, R);
  push_up(rt);
}
int main() {
  int n, q;
  cin >> n >> q;
  init();
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, n, 1);
  while (q--) {
    int op, l, r;
    scanf("%d %d %d", &op, &l, &r);
    if (op == 1) {
      update(1, n, 1, l, r);
    } else {
      cout << query(1, n, 1, l, r) << "\n";
    }
  }
}