Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.

There are n stages available. The rocket must contain exactly k of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.

For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — 26 tons.

Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.

Input

The first line of input contains two integers — n and k (1 ≤ k ≤ n ≤ 50) – the number of available stages and the number of stages to use in the rocket.

The second line contains string s, which consists of exactly n lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.

Output

Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.

Examples

Input

5 3
xyabd


Output

29

Input

7 4
problem


Output

34

Input

2 2
ab


Output

-1

Input

12 1
abaabbaaabbb


Output

1

Note

In the first example, the following rockets satisfy the condition:

  * "adx" (weight is 1+4+24=29);
  * "ady" (weight is 1+4+25=30);
  * "bdx" (weight is 2+4+24=30);
  * "bdy" (weight is 2+4+25=31).



Rocket "adx" has the minimal weight, so the answer is 29.

In the second example, target rocket is "belo". Its weight is 2+5+12+15=34.

In the third example, n=k=2, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;


 
public class Try
{
    
    static class InputReader {
 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
 
		public InputReader(InputStream st) {
			this.stream = st;
		}
 
		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
 
		public int ni() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nl() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nia(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
 
		public String rs() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
 
		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}
 
		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
 
	}
	

	
	
	
    static long mod=1000000007;
    static BigInteger bigInteger = new BigInteger("1000000007");
    static int n = (int)1e6;
    static boolean[] prime;
    static ArrayList<Integer> as;
    static void sieve() {
			Arrays.fill(prime	, true);
			prime[0] = prime[1] = false;
			for(int i = 2 ; i * i <= n ; ++i) {
				if(prime[i]) {
					for(int k = i * i; k< n ; k+=i) {
						prime[k] = false;
					}
				}
			}
			
		}
		static PrintWriter w = new PrintWriter(System.out);
 static char [][]sol;
    public static void main(String[] args)
    {
        
        InputReader sc = new InputReader(System.in);
        //PrintWriter w = new PrintWriter(System.out);
    
       /* 
        prime  = new boolean[n + 1];
		sieve();
		prime[1] = false;
		*/
		/*
		as = new ArrayList<>();
		for(int i=2;i<=1000000;i++)
		{
		    if(prime[i])
		    as.add(i);
		}
		*/
		
       /* 
		
	long a = sc.nl();
	 
            BigInteger ans = new BigInteger("1");
            for (long i = 1; i < Math.sqrt(a); i++) {
                if (a % i == 0) {
                    if (a / i == i) {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        
                    } else {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        ans = ans.multiply(BigInteger.valueOf(phi(a / i)));
 
                        
                    }
 
 
                }
            }
            w.println(ans.mod(bigInteger));
            */
          
         // MergeSort ob = new MergeSort();
       // ob.sort(arr, 0, arr.length-1);
       
    // Student []st = new Student[x];    
   // st[i] = new Student(i,d[i]);
     //Arrays.sort(st,(p,q)->p.diff-q.diff);
    
     
    int x = sc.ni();
    int k = sc.ni();
    int []a = new int[26];
    String s = sc.rs();
    for(int i=0;i<x;i++)
    {
        a[(int) s.charAt(i) -97]++;
    }
    //HashMap<Integer,Integer> hm = new HashMap
    int res = 0;
    int curr = -2;
    int p = 0;
    for(int i=0;i<26;i++)
    {
        
        if(a[i] > 0 && i - curr > 1)
        {
            curr = i;
            res += (i+1);
            p++;
            if(p == k)
                break;
        }
        
    }
     
     if(p == k)
     w.println(res);
     else
         w.println("-1");
     
     
     
     
     
     
     
   
   
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
     
    
      w.close();
        
    
    
    
    
    
   
  
    
    }
    static int modexp(int x,int n,int M)
{
    int power = n;
    int result=1;
    while(power>0)
    {
        if(power % 2 ==1)
            result=(result * x)%M;
        x=(x*x)%M;
        power = power/2;
    }
    return result;
}
   
     static class Student
    {
        
        long first;
        long sec;
        Student(long first,long sec)
        {
            this.first = first;
            this.sec = sec;
            
            
        }
        
    }
     
     public static long modMultiply(long one, long two) {
 
		return (one % mod * two % mod) % mod;
	}
    static long fx(int m)
    {
        long re = 0;
        for(int i=1;i<=m;i++)
        {
            re += (long) (i / gcd(i,m));
        }
        return re;
    }
     static long gcd(long a, long b)
    {
	 if (a == 0)
	  return b;
				         
	return gcd(b%a, a);
    }
    
   
    
     static long phi(long nx)
    {
        // Initialize result as n
        double result = nx; 
        
     
        // Consider all prime factors of n and for 
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 0; as.get(p) * as.get(p) <= nx; ++p)
        {
            // Check if p is a prime factor.
            if (nx % as.get(p) == 0)
            {
                // If yes, then update n and result
                while (nx % as.get(p) == 0)
                    nx /= as.get(p);
                result *= (1.0 - (1.0 / (double) as.get(p)));
            }
        }
     
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (nx > 1)
            result *= (1.0 - (1.0 / (double) nx));
     
        return (long)result;
    
        
        //return(phi((long)result,k-1));
        
    }
    
    public static int primeFactors(int n,int x)
    {
        as = new ArrayList<>();
        int sum = 0;
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            
            if(sum == x-1)
                break;
            as.add(2);
            //System.out.print(2 + " ");
            n /= 2;
            sum++;
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
            
               // System.out.print(i + " ");
                if(sum == x-1)
                break;
                as.add(i);
                n /= i;
                sum++;
                
            }
            
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n >= 2)
        {
            sum++;
            as.add(n);
        }
            return sum;
    }
    static int digitsum(int x)
    
    {    
 
        int sum = 0;
        while(x > 0)
        {
            int temp = x % 10;
            sum += temp;
            x /= 10;
        }
        return sum;
        
    }
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i*i <=n; i++)
        {
             if (n % i == 0 && i<=1000000)
             {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                   cnt++;
      
                else // Otherwise count both
                    cnt = cnt + 2;
             }
         }
        return cnt;
    }
   
    static boolean isprime(int n)
    {
    
    if(n == 2)
        return true;
    if(n == 3)
        return true;
    if(n % 2 == 0)
        return false;
    if(n % 3 == 0)
        return false;
 
    int i = 5;
   int w = 2;
 
    while(i * i <= n)
    {
        if(n % i == 0)
            return false;
    
 
        i += w;
        w = 6 - w;
    }
 
    return true;
    }
    
    static long log2(long value) {
    return Long.SIZE-Long.numberOfLeadingZeros(value);
}
 
 
 
 
    
    
    static boolean binarysearch(ArrayList<Integer> arr,int p,int n)
    {
        //ArrayList<Integer> as = new ArrayList<>();
        //as.addAll(0,at);
        //Collections.sort(as);
        boolean re = false;
        int st = 0;
        int end = n-1;
        
        
        
        
        while(st <= end)
        { 
            int mid = st + (end-st)/2;
        
            if(p > arr.get(mid))
            {
                st = mid+1;
                
            }
            else if(p < arr.get(mid))
            {
                end = mid-1;
                
            }
            else if(p == arr.get(mid))
            {
            
            re = true;
            break;
            }
           
            
            
        }
         return re;
        
    }
    
   
    
    /* Java program for Merge Sort */
static class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
 
    /* A utility function to print array of size n */
    
 
   
}
    

    
    
    
    public static int ip(String s){
		return Integer.parseInt(s);
	}
    
	 public static String ips(int s){
		return Integer.toString(s);
	}
	
    
					 
}