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 N = 3e5 + 5;
const int M = 1e6;
long long n, m, maxn[N << 2], sum[N << 2], d[M], num[N];
long long read() {
  long long res, f = 1;
  char ch;
  while ((ch = getchar()) < '0' || ch > '9')
    if (ch == '-') f = -1;
  res = ch - 48;
  while ((ch = getchar()) >= '0' && ch <= '9') res = res * 10 + ch - 48;
  return res * f;
}
long long max(long long x, long long y) {
  if (x > y) return x;
  return y;
}
void pushup(long long k) {
  sum[k] = sum[k << 1] + sum[k << 1 | 1];
  maxn[k] = max(maxn[k << 1], maxn[k << 1 | 1]);
}
void build(long long k, long long l, long long r) {
  if (l == r) {
    sum[k] = maxn[k] = num[l];
    return;
  }
  long long mid = l + r >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
  pushup(k);
}
void update(long long k, long long l, long long r, long long x, long long y) {
  if (maxn[k] <= 2) return;
  if (l == r) {
    num[l] = d[num[l]];
    sum[k] = maxn[k] = num[l];
    return;
  }
  long long mid = l + r >> 1;
  if (x <= mid) update(k << 1, l, mid, x, y);
  if (mid < y) update(k << 1 | 1, mid + 1, r, x, y);
  pushup(k);
}
long long query(long long k, long long l, long long r, long long x,
                long long y) {
  if (l >= x && r <= y) return sum[k];
  long long mid = l + r >> 1, ans = 0;
  if (x <= mid) ans += query(k << 1, l, mid, x, y);
  if (mid < y) ans += query(k << 1 | 1, mid + 1, r, x, y);
  return ans;
}
void pre() {
  long long i, j;
  for (i = 2; i <= M; i++) {
    for (j = i; j <= M; j += i) d[j]++;
    d[i]++;
  }
}
int main() {
  long long i, x, y, op;
  n = read();
  m = read();
  for (i = 1; i <= n; i++) num[i] = read();
  pre();
  build(1, 1, n);
  while (m--) {
    op = read();
    x = read();
    y = read();
    if (op == 1) update(1, 1, n, x, y);
    if (op == 2) printf("%lld\n", query(1, 1, n, x, y));
  }
  return 0;
}