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 n, m;
long long a[300005];
long long t, l, r;
long long sumt[1200005], maxt[1200005];
long long D[1000005];
void build(long long u, long long L, long long R) {
  if (L == R) {
    maxt[u] = sumt[u] = a[L];
  } else {
    long long M = (R + L) >> 1;
    build(u * 2, L, M);
    build(u * 2 + 1, M + 1, R);
    maxt[u] = max(maxt[u * 2], maxt[u * 2 + 1]);
    sumt[u] = sumt[u * 2] + sumt[u * 2 + 1];
  }
}
long long getsum(long long u, long long L, long long R, long long l,
                 long long r) {
  long long ret = 0;
  if (L == l && R == r) {
    ret = sumt[u];
  } else {
    long long M = (R + L) >> 1;
    if (l <= M) {
      ret += getsum(u * 2, L, M, l, min(r, M));
    }
    if (r >= M + 1) {
      ret += getsum(u * 2 + 1, M + 1, R, max(M + 1, l), r);
    }
  }
  return ret;
}
void upd(long long u, long long L, long long R, long long l, long long r) {
  if (maxt[u] <= 2) {
    return;
  }
  if (L == R) {
    sumt[u] = D[sumt[u]];
    maxt[u] = sumt[u];
  } else {
    long long M = (R + L) >> 1;
    if (l <= M) {
      upd(u * 2, L, M, l, min(r, M));
    }
    if (r >= M + 1) {
      upd(u * 2 + 1, M + 1, R, max(M + 1, l), r);
    }
    maxt[u] = max(maxt[u * 2], maxt[u * 2 + 1]);
    sumt[u] = sumt[u * 2] + sumt[u * 2 + 1];
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 1; i <= 1000000; ++i) {
    for (long long j = i; j <= 1000000; j += i) {
      ++D[j];
    }
  }
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build(1, 1, n);
  for (long long j = 1; j <= m; ++j) {
    cin >> t >> l >> r;
    if (t == 1) {
      upd(1, 1, n, l, r);
    } else {
      cout << getsum(1, 1, n, l, r) << '\n';
    }
  }
}