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:
/* Idea is that after at most 1o replace operations a number will become contstant. After that we do not need to aply replce 
operations. Use dsu to find next index where we can apply replace operation. 
Use BIT to update and find sum in given range
*/
import java.util.*;
import java.io.*;
import java.math.BigInteger; 
public class lp{ 
static PrintWriter out = new PrintWriter(System.out);

static void fun(int div[])
{
       for(int i=1;i<div.length;i++)
          for(int j=i;j<div.length;j=j+i)
              div[j]++;
}
static void update(long BIT[],int i,int sum)
{
     while(i<BIT.length)
       {
           BIT[i] = BIT[i]+sum;
               i = i + (i&(-i));        
        }
}
static long sum(long BIT[],int i)
{
     long s=0;
    while(i>0)
      {
             s=s+BIT[i];
             i = i - (i&(-i));
      }
   return s;
}
static int find(int x,int p[])
{
     if(x==p[x])
        return x;
      return p[x] = find(p[x],p);
}

static void update(int l,int r,long bit[],int p[],int a[],int ct[],int div[],int n)
{
      for(int i=l;i<=r;)
        {
            int root = find(i,p);
            if(root>r)
               return;
            update(bit,root+1,div[a[root]]-a[root]);
            a[root] = div[a[root]];
            ct[root]++;
            if(ct[root]==10)
              {
                 int rr = find(root+1,p);                
                 p[root]=rr;
               }
            i=root+1;
        }        
}
static void solve() {

int div[] = new int[1000002];
fun(div);
int n = ni();
int m = ni();
int a[] = ai(n);
int ct[] = new int[n];
int p[] = new int[n+1];
long bit[] = new long[n+1];
for(int i=0;i<n;i++){
     update(bit,i+1,a[i]);
     p[i]=i;
}
 
     p[n]=n;
for(int o=1;o<=m;o++)
{

int t = ni();
int l=ni();
int r = ni();
if(t==1)
 update(l-1,r-1,bit,p,a,ct,div,n);
else{
long s = sum(bit,r)-sum(bit,l-1);
  pn(s);
}
}
 
   out.flush();
}
public static void main(String[] args) throws Exception{


// use this block when you need more recursion depth
new Thread(null, null, "Name", 999999999) {
            public void run() {
                try {
                    solve();
                } catch(Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();  
} 
static int[] ai(int n)  // it will give in array of size n
{
    int a[] = new int[n];
    for(int i=0;i<n;i++)
        a[i] = ni();
    return a;
}
static long[] al(int n)  // it will give in array of size n
{
    long a[] = new long[n];
    for(int i=0;i<n;i++)
        a[i] = nl();
    return a;
}
 
static void p(Object o)
{
   out.print(o);
}
static void pn(Object o)
{
    out.println(o);
}
static int abs(int x)
{
    return x>0 ? x : -x;
}
static long gcd(long a,long b)
{
       if(a==0)
          return b;
       if(b==0)
          return a;
       if(b%a==0)
          return a;
          return gcd(b%a,a);
}
static int count_set(int n)
{
   int c=0;
while(n>0)
{
   if(n%2==1)
     c++;
     n=n/2;
}
   return c;
}
static void subtract_1(char s[]) // it will subtract 1 from the given number.  number should be positive
{
      if(s[0]=='0') // number is zero
          return;
     int n = s.length,i=n-1;
     while(s[i]=='0')
       i--;
     s[i] = (char)((int)(s[i]-'0') + 47);
   
     for(int j=i+1;j<n;j++)
         s[j]='9';
}
static long pow(long a,long b,long md)
{ 
    long ans=1;
    while(b>0)
    {
         if(b%2==1)
           ans = (ans*a)%md;
           a = (a*a)%md;
           b = b/2;
    }
   return ans;
}
static long min(long a,long b){
 
     return a<b ? a : b;
}
static long max(long a,long b){
     return a>b ? a : b;
}
static boolean pal(String s)
{
   int n = s.length(),i1=0,i2=n-1;
 
     while(i1<i2)
     {
        if(s.charAt(i1)!=s.charAt(i2))
             return false;
             i1++; i2--;
     }
       return true;
}
static String rev(String r)
{
    String s = "";
    int i= r.length()-1;
 
    while(i>=0)
    {
        s=s+r.charAt(i);
        i--;   
    }
       return s;
}
static FastReader sc=new FastReader(); 
 
      static int ni(){
                 int x = sc.nextInt();
                 return(x);
        }
      static long nl(){
              long x = sc.nextLong();
              return(x);
         }
      static String n(){
                 String str = sc.next();
                     return(str);
       }
     static String ns(){
                 String str = sc.nextLine();
                   return(str);
      }
     static double nd(){
               double d = sc.nextDouble();
                 return(d);
       }
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
}