Problem: 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

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
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
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastScanner in = new FastScanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ecr61F solver = new ecr61F();
        solver.solve(1, in, out);
        out.close();
    }

    static class ecr61F {
        public void solve(int testNumber, FastScanner in, PrintWriter out) {
            int n = in.ni();
            String s = in.ns();
            int[][] dp = new int[n][n];
            for (int i = 0; i < n; i++) {
                dp[i][i] = 1;
            }
            for (int len = 2; len <= n; len++) {
                for (int i = 0; i <= n - len; i++) {
                    int best = len + 7;
                    for (int k = i; k < i + len - 1; k++) {
                        if (s.charAt(k) == s.charAt(i + len - 1)) {
                            best = Math.min(best, dp[i][k] + dp[k + 1][i + len - 1] - 1);
                        } else {
                            best = Math.min(best, dp[i][k] + dp[k + 1][i + len - 1]);
                        }
                    }
                    dp[i][i + len - 1] = best;
                }
            }
            out.println(dp[0][n - 1]);

        }

    }

    static class FastScanner {
        private BufferedReader in;
        private StringTokenizer st;

        public FastScanner(InputStream stream) {
            in = new BufferedReader(new InputStreamReader(stream));
        }

        public String ns() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    String rl = in.readLine();
                    if (rl == null) {
                        return null;
                    }
                    st = new StringTokenizer(rl);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(ns());
        }

    }
}