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 JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeSet;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskF {
        final int N = (int) 1e6 + 10;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int[] a = new int[n + 1];
            TreeSet<Integer> set = new TreeSet<>();
            BIT bit = new BIT(n + 1);
            for (int i = 1; i <= n; ++i) {
                int x = in.nextInt();
                a[i] = x;
                if (x > 2) set.add(i);
                bit.add(i, x);
            }
            int[] D = new int[N];
            for (int i = 1; i < N; ++i) {
                for (int j = i; j < N; j += i) {
                    D[j]++;
                }
            }
            for (int q = 0; q < m; ++q) {
                int type = in.nextInt();
                int l = in.nextInt();
                int r = in.nextInt();
                if (type == 1) {
                    for (Integer i = set.ceiling(l); i != null && i <= r; i = set.higher(i)) {
                        int di = D[a[i]];
                        bit.add(i, -a[i]);
                        a[i] = di;
                        bit.add(i, a[i]);
                        if (di <= 2) set.remove(i);
                    }
                } else {
                    long sum = bit.sum(l, r);
                    out.println(sum);
                }
            }
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }

    static class BIT {
        long[] tree;
        int n;

        public BIT(int n) {
            this.n = n;
            tree = new long[n + 1];
        }

        public void add(int i, long val) {
            while (i <= n) {
                tree[i] += val;
                i += i & -i;
            }
        }

        public long sum(int to) {
            long res = 0;
            for (int i = to; i >= 1; i -= (i & -i)) {
                res += tree[i];
            }
            return res;
        }

        public long sum(int from, int to) {
            return sum(to) - sum(from - 1);
        }

    }
}