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 int ara[300007], divisor[4 * 300007], tree[4 * 300007],
    mxTree[4 * 300007];
void InitiaLization() {
  for (long long int i = 1; i <= 1000000; i++)
    for (long long int j = i; j <= 1000000; j += i) divisor[j]++;
}
void build(long long int node, long long int b, long long int e) {
  if (b == e) {
    tree[node] = ara[b];
    mxTree[node] = ara[b];
    return;
  }
  long long int mid = (b + e) / 2, lson = 2 * node, rson = lson + 1;
  build(lson, b, mid);
  build(rson, mid + 1, e);
  tree[node] = tree[lson] + tree[rson];
  mxTree[node] = max(mxTree[lson], mxTree[rson]);
}
void update(long long int node, long long int b, long long int e,
            long long int x, long long int y) {
  if (b > y || e < x) return;
  if (mxTree[node] <= 2) return;
  if (b == e) {
    tree[node] = divisor[tree[node]];
    mxTree[node] = tree[node];
    return;
  }
  long long int mid = (b + e) / 2, lson = 2 * node, rson = lson + 1;
  update(lson, b, mid, x, y);
  update(rson, mid + 1, e, x, y);
  tree[node] = tree[lson] + tree[rson];
  mxTree[node] = max(mxTree[lson], mxTree[rson]);
}
long long int query(long long int node, long long int b, long long int e,
                    long long int x, long long int y) {
  if (b > y || e < x) return 0;
  if (b >= x && e <= y) return tree[node];
  long long int mid = (b + e) / 2, lson = 2 * node, rson = lson + 1;
  return query(lson, b, mid, x, y) + query(rson, mid + 1, e, x, y);
}
int main(int argc, char const *argv[]) {
  InitiaLization();
  long long int n, m;
  scanf("%lld %lld", &n, &m);
  for (long long int i = 1; i <= n; i++) scanf("%d", ara + i);
  build(1, 1, n);
  while (m--) {
    long long int t, l, r;
    scanf("%lld %lld %lld", &t, &l, &r);
    if (t == 1) {
      update(1, 1, n, l, r);
    } else {
      printf("%lld\n", query(1, 1, n, l, r));
    }
  }
  return 0;
}