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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Liavontsi Brechka
 */
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);
        FUdaliStroku solver = new FUdaliStroku();
        solver.solve(1, in, out);
        out.close();
    }

    static class FUdaliStroku {
        InputReader in;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            this.in = in;

            int n = ni();
            String s = next();

            int[][] dp = new int[n][n];
            for (int i = 0; i < n; i++) {
                Arrays.fill(dp[i], Integer.MAX_VALUE >>> 1);
                dp[i][i] = 1;
            }

            for (int length = 2; length <= n; length++) {
                for (int l = 0; l + length - 1 < n; l++) {
                    int r = l + length - 1;
                    dp[l][r] = Math.min(dp[l][r], 1 + dp[l + 1][r]);

                    for (int i = l + 1; i <= r; i++) {
                        if (s.charAt(l) == s.charAt(i)) {
                            dp[l][r] = Math.min(dp[l][r], (l + 1 <= i - 1 ? dp[l + 1][i - 1] : 0) + dp[i][r]);
                        }
                    }
                }
            }

            out.print(dp[0][n - 1]);
        }

        private int ni() {
            return in.nextInt();
        }

        private String next() {
            return in.next();
        }

    }

    static class InputReader {
        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
        }

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

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(readLine());
            }
            return tokenizer.nextToken();
        }

        public String readLine() {
            String line;
            try {
                line = reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

    }
}

