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;
const long long N = 3e5 + 1;
const long long M = 1e6 + 1;
long long sum[N << 2], MAX[N << 2], num[M];
void push(long long rt) {
  MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]);
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void updbuild(long long l, long long r, long long rt) {
  if (l == r) {
    scanf("%lld", &sum[rt]);
    MAX[rt] = sum[rt];
    return;
  }
  long long m = (l + r) >> 1;
  updbuild(l, m, rt << 1);
  updbuild(m + 1, r, rt << 1 | 1);
  push(rt);
}
void reupd(long long L, long long R, long long l, long long r, long long rt) {
  if (l == r) {
    MAX[rt] = num[MAX[rt]];
    sum[rt] = num[sum[rt]];
    return;
  }
  long long m = (l + r) >> 1;
  if (L <= m && MAX[rt << 1] > 2) reupd(L, R, l, m, rt << 1);
  if (R > m && MAX[rt << 1 | 1] > 2) reupd(L, R, m + 1, r, rt << 1 | 1);
  push(rt);
}
long long query(long long L, long long R, long long l, long long r,
                long long rt) {
  if (L <= l && r <= R) return sum[rt];
  long long m = (l + r) >> 1;
  long long ans = 0;
  if (L <= m) ans += query(L, R, l, m, rt << 1);
  if (R > m) ans += query(L, R, m + 1, r, rt << 1 | 1);
  return ans;
}
void init() {
  for (long long i = 1; i < M; i++)
    for (long long j = i; j < M; j += i) num[j]++;
}
int main() {
  init();
  long long n, m, op, a, b;
  scanf("%lld%lld", &n, &m);
  updbuild(1, n, 1);
  while (m--) {
    scanf("%lld%lld%lld", &op, &a, &b);
    if (op == 1)
      reupd(a, b, 1, n, 1);
    else
      printf("%lld\n", query(a, b, 1, n, 1));
  }
  return 0;
}
