One unknown hacker wants to get the admin's password of AtForces testing system, to get problems from the next contest. To achieve that, he sneaked into the administrator's office and stole a piece of paper with a list of n passwords — strings, consists of small Latin letters.

Hacker went home and started preparing to hack AtForces. He found that the system contains only passwords from the stolen list and that the system determines the equivalence of the passwords a and b as follows:

  * two passwords a and b are equivalent if there is a letter, that exists in both a and b; 
  * two passwords a and b are equivalent if there is a password c from the list, which is equivalent to both a and b. 



If a password is set in the system and an equivalent one is applied to access the system, then the user is accessed into the system.

For example, if the list contain passwords "a", "b", "ab", "d", then passwords "a", "b", "ab" are equivalent to each other, but the password "d" is not equivalent to any other password from list. In other words, if:

  * admin's password is "b", then you can access to system by using any of this passwords: "a", "b", "ab"; 
  * admin's password is "d", then you can access to system by using only "d". 



Only one password from the list is the admin's password from the testing system. Help hacker to calculate the minimal number of passwords, required to guaranteed access to the system. Keep in mind that the hacker does not know which password is set in the system.

Input

The first line contain integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of passwords in the list. Next n lines contains passwords from the list – non-empty strings s_i, with length at most 50 letters. Some of the passwords may be equal.

It is guaranteed that the total length of all passwords does not exceed 10^6 letters. All of them consist only of lowercase Latin letters.

Output

In a single line print the minimal number of passwords, the use of which will allow guaranteed to access the system.

Examples

Input


4
a
b
ab
d


Output


2

Input


3
ab
bc
abc


Output


1

Input


1
codeforces


Output


1

Note

In the second example hacker need to use any of the passwords to access the system.
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.*; 
import java.util.StringTokenizer; 
import java.io.PrintWriter;
import java.io.*;
import java.util.stream.Collectors.*;
import java.lang.*;
import static java.util.stream.Collectors.*;
import static java.util.Map.Entry.*;
public class Ideo
{ 
    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; 
        } 
    }
   
static int power(int x,int y) 
{ 
    int res = 1;      // Initialize result 
  
      // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y%2==1) 
            res = (res*x); 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x);   
    } 
    return res; 
} 
 
 static int gcd(int a, int b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    } 


  static boolean compareSeq(char[] S, int x, int y, int n) 
    { 
        for (int i = 0; i < n; i++)
         {

            if (S[x] < S[y]) 
               return true;
            else if (S[x] > S[y]) 
                return false; 
            x = (x + 1) % n; 
            y = (y + 1) % n; 
        } 
        return true; 
    } 

    static void build(long[] sum,int[] arr,int n)
    {
      for(int i=0;i<(1<<n);i++)
      {
        long total=0;
        for(int j=0;j<n;j++)
        {
          if((i & (1 << j)) > 0)
            total+=arr[j];
        }

        sum[i]=total;
      }
    }

 
     static int count(long arr[], long x, int n) 
       {

        int l=0;
        int h=n-1;

        int res=-1;
        int mid=-1;
        while(l<=h)
        {
          mid=l+(h-l)/2;
          if(x==arr[mid])
          {
            res=mid;
            h=mid-1;
          }
          else if(x<arr[mid])
            h=mid-1;
          else
            l=mid+1;
        }

        if(res==-1)
          return 0;

        //res is first index and res1 is last index of an element in a sorted array total number of occurences is (res1-res+1)

        int res1=-1;
        l=0;
        h=n-1;
        while(l<=h)
        {
          mid=l+(h-l)/2;
          if(x==arr[mid])
          {
            res1=mid;
            l=mid+1;
          }
          else if(x<arr[mid])
            h=mid-1;
          else
            l=mid+1;
        }



        if(res1==-1)
          return 0;
        if(res!=-1 && res1!=-1)
          return (res1-res+1);

        return 0;

       } 

       static int parity(int a)
       {
    a^=a>>16;
    a^=a>>8;
    a^=a>>4;
    a^=a>>2;
    a^=a>>1;
    return a&1;
  }
/*
  PriorityQueue<aksh> pq = new PriorityQueue<>((o1, o2) -> {
            if (o1.p < o2.p)
                return 1;
            else if (o1.p > o2.p)
                return -1;
            else
                return 0;

        });//decreasing order acc to p*/
static int power(int x, int y, int m)  
    { 
        if (y == 0) 
            return 1; 
          
        int p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    } 
static int modinv(int a, int m) 
    { 
        int g = gcd(a, m); 
        if (g != 1) 
            return 0;
        else 
        { 
           return power(a, m - 2, m); 
        } 
        //return 0;
    } 

static int[] product(int[] nums) {
    int[] result = new int[nums.length];
 
    int[] t1 = new int[nums.length];
    int[] t2 = new int[nums.length];
 
    t1[0]=1;
    t2[nums.length-1]=1;
 
    //scan from left to right
    for(int i=0; i<nums.length-1; i++){
        t1[i+1] = nums[i] * t1[i];
    }
 
    //scan from right to left
    for(int i=nums.length-1; i>0; i--){
        t2[i-1] = t2[i] * nums[i];
    }

    for(int i=0;i<nums.length;i++)
      {
        System.out.print(t1[i]+" "+t2[i]);
        System.out.println();
      }
 
    //multiply
    for(int i=0; i<nums.length; i++){
        result[i] = t1[i] * t2[i];
    }
 
    return result;
}

static int getsum(int[] bit,int ind)
{
  int sum=0;
  while(ind>0)
  {
    sum+=bit[ind];
    ind-= ind & (-ind);
  }

  return sum;
}

static void update(int[] bit,int max,int ind,int val)
{
  while(ind<=max)
  {
    bit[ind]+=val;
    ind+= ind & (-ind);
  }
}

    //static ArrayList<Integer>[] adj;
    static int[] bfs(ArrayList<Integer>[] adj,int n,ArrayList<Integer> arr)
    {
      int[] vis=new int[n];
      int[] dis=new int[n];
      int[] par=new int[n];
      Queue<Integer> q=new LinkedList<>();

      Arrays.fill(dis,-1);

      for(int i=0;i<arr.size();i++)
      {
        vis[arr.get(i)]=1;
        dis[arr.get(i)]=0;
        par[arr.get(i)]=-1;
        q.add(arr.get(i));
      }

      while(!q.isEmpty())
      {
        int x=q.remove();
        for(int u:adj[x])
        {
          if(vis[u]==0)
          {
            vis[u]=1;
            dis[u]=dis[x]+1;
            par[u]=x;
            q.add(u);
          }
        }
      }

      return dis;
    }

    static boolean check(long mid,long a,long b)
    {
      long count=1;
      while(count<=mid)
      {
        count++;
        if(a<b)
          a+=count;
        else
          b+=count;

        if(a==b)
          return true;
      }
        return false;
    }

    static class a1 implements Comparable<a1>
    {
      int p;
      int q;

      public a1(int p,int q)
      {
        this.p=p;
        this.q=q;
      }

      public int compareTo(a1 o)
      {
        return p-o.p;
      }
    }
    static int get(int arr[], int n) 
    { 
        int result = 0; 
        int x, sum; 
          
        // Iterate through every bit 
        for(int i=0; i<32; i++) 
        { 
            // Find sum of set bits at ith position in all  
            // array elements 
            sum = 0; 
            x = (1 << i); 
            for(int j=0; j<n; j++) 
            { 
                if((arr[j] & x)!=0) 
                sum++; 
            } 
            // The bits with sum not multiple of 3, are the 
            // bits of element with single occurrence. 
            if ((sum % 3)!=0) 
            result |= x; 
        } 
        return result; 
    } 
    /*        Collections.sort(orders, new Comparator<Point>() {
            @Override
            public int compare(Point o1, Point o2) {
                if (o1.diff < o2.diff) {
                    return -1;
                } else if (o1.diff > o2.diff) {
                    return 1;
                } else {
                    return 0;
                }
            }
        });  */


    static boolean isPrime(int n)  
    {  
        // Corner cases  
        if (n <= 1) return false;  
        if (n <= 3) return true;  
          
        // This is checked so that we can skip  
        // middle five numbers in below loop  
        if (n % 2 == 0 || n % 3 == 0) return false;  
          
        for (int i = 5; i * i <= n; i = i + 6)  
            if (n % i == 0 || n % (i + 2) == 0)  
            return false;  
          
        return true;  
    }  
      
    static int nextPrime(int N)  
    {  
      
        // Base case  
        if (N <= 1)  
            return (2-N);  
      
        int prime = N;  
        boolean found = false;  
      
        // Loop continuously until isPrime returns  
        // true for a number greater than n  
        while (!found)  
        {  
            if (isPrime(prime))  
                {
                  found = true;
                  break;
                }  

            prime++;
        }  
      
        return (prime-N);  
    }  


    static long product(long x) 
{ 
    long prod = 1; 
    while (x > 0)  
    { 
        prod *= (x % 10); 
        x /= 10; 
    } 
    return prod; 
} 
  
// This function returns the number having 
// maximum product of the digits 
static long findNumber(long l, long r) 
{ 
    // Converting both integers to strings 
    //string a = l.ToString(); 
    String b = Long.toString(r); 
  
    // Let the current answer be r 
    long ans = r; 
    for (int i = 0; i < b.length(); i++)  
    { 
        if (b.charAt(i) == '0') 
            continue; 
  
        // Stores the current number having 
        // current digit one less than current 
        // digit in b 
        char[] curr = b.toCharArray(); 
        curr[i] = (char)(((int)(curr[i] -  
                    (int)'0') - 1) + (int)('0')); 
  
        // Replace all following digits with 9 
        // to maximise the product 
        for (int j = i + 1; j < curr.length; j++) 
            curr[j] = '9'; 
  
        // Convert string to number 
        int num = 0; 
        for (int j = 0; j < curr.length; j++) 
            num = num * 10 + (curr[j] - '0'); 
  
        // Check if it lies in range and its product 
        // is greater than max product 
        if (num >= l && product(ans) < product(num)) 
            ans = num; 
    } 
  
    return product(ans); 
} 

static long mod=998244353;

static long pow(long in, long pow) {
    if(pow == 0) return 1;
    long out = pow(in, pow / 2);
    out = (out * out) % mod;
    if(pow % 2 == 1) out = (out * in) % mod;
    return out;
  }
  static long inv(long in) {
    return pow(in, mod - 2);
  }

  static void swap(int x,int y)
  {
    int temp=x;
    x=y;
    y=temp;
  }

  static int[] par;
  static int[] size;
  static int find(int i) 
  {
     if (par[i] == i)
      return i;
 
    return par[i] = find(par[par[i]]);
  }
 
  static void union(int x, int y) 
  {
      x = find(x);
      y = find(y);
      if (x == y)
        return;
      if (size[x] < size[y])
        swap(x, y);
 
      par[y] = x;
      size[x] += size[y];
    }

    static final long INF = Long.MAX_VALUE/5;
    static ArrayList<Integer>[] adj;
    public static void main(String args[] ) throws Exception 
    {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT */

        FastReader sc=new FastReader();
       

        int n=sc.nextInt();
        String[] s=new String[n];

        par=new int[27];
        size=new int[27];
        int[] vis=new int[27];
        for(int i=1;i<=26;i++)
          {
            par[i]=i;
            size[i]=1;
          }


        for(int i=0;i<n;i++)
        {
          s[i]=sc.next();
          for(int j=0;j<s[i].length();j++)
            vis[s[i].charAt(j)-'a'+1]=1;
        }

        int res=0;
        for(int i=0;i<n;i++)
        {
          for(int j=0;j<s[i].length()-1;j++)
          {
            int a = s[i].charAt(j)-'a'+1;
            int b = s[i].charAt(j+1)-'a'+1;
            union(a,b);
          }
        }

        for(int i=1;i<=26;i++)
        {
          if(par[i]==i && vis[i]==1)
            res++;
        }

        System.out.println(res);
        
        
      }

    }


