You are given a string s of length n consisting of lowercase Latin letters. You may apply some operations to this string: in one operation you can delete some contiguous substring of this string, if all letters in the substring you delete are equal. For example, after deleting substring bbbb from string abbbbaccdd we get the string aaccdd.

Calculate the minimum number of operations to delete the whole string s.

Input

The first line contains one integer n (1 ≤ n ≤ 500) — the length of string s.

The second line contains the string s (|s| = n) consisting of lowercase Latin letters.

Output

Output a single integer — the minimal number of operation to delete string s.

Examples

Input


5
abaca


Output


3

Input


8
abcddcba


Output


4
import java.util.*;
import java.io.*;

public class ClearTheString {
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int N = scanner.nextInt();
        String S = scanner.next();
        String comp = S.substring(0,1);
        for(int i = 1; i < N; i++) {
            if (S.charAt(i)!= S.charAt(i-1)) comp+=S.charAt(i);
        }
        int cur = comp.length();
        boolean[][] good = new boolean[cur][cur];
        for(int i = 0; i < cur; i++) {
            for(int j = 0; j < cur; j++) {
                if (comp.charAt(i) == comp.charAt(j)) good[i][j] = true;
            }
        }
        //can also do the dp with ijk but this is more intuitive
        int[][] dp = new int[cur][cur];
        //length of interval
        for(int len = 1; len < cur; len++) {
            //first index
            for(int i = 0; i < cur - len; i++) {
                //second index
                int j = i + len;
                //match at this index
                if (good[i][j]) {
                    //Then can remove by removing inner and single removal for both i and j
                    dp[i][j] = dp[i+1][j-1]+1;
                    //check all other valid points in interval where we match
                    for(int k = i+2; k < j-1; k++) {
                        if (good[i][k] && good[k][j]) {
                            //todo not 100% sure why this works :/
                            dp[i][j] = Math.max(dp[i][j], dp[i][k] + dp[k][j]);
                        }
                    }
                }
                //check all internal points for better mtaching
                for(int k = i; k < j; k++) {
                    dp[i][j] = Math.max(dp[i][j], dp[i][k] + dp[k+1][j]);
                }
                
            }
        }
        out.println(cur-dp[0][cur-1]);
        out.flush();
    }
    
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(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 readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}
