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 MainClass
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());
        char[] S = in.readLine().toCharArray();
        char[] A = new char[S.length];
        int c = 0;
        for (int i=0;i<n;i++)
        {
            int j = i;
            while (j < n && S[j] == S[i])   j++;
            A[c++] = S[i];
            i = j - 1;
        }
        n = c;
        int[][] dp = new int[n][n];
        for (int i=n - 1;i>=0;i--)
        {
            for (int j=i;j<n;j++)
            {
                if (i == j) dp[i][j] = 1;
                else
                {
                    if (j - i == 1)
                    {
                        if (A[i] == A[j])   dp[i][j] = 1;
                        else    dp[i][j] = 2;
                    }
                    else
                    {
                        dp[i][j] = dp[i + 1][j] + 1;
                        for (int k=i + 1;k<=j;k++)
                        {
                            if (A[k] == A[i])
                            {
                                int x = dp[k][j];
                                if (k - 1 >= i + 1)
                                    x += dp[i + 1][k - 1];
                                dp[i][j] = Math.min(dp[i][j], x);
                            }
                        }
                    }
                }
            }
        }
        System.out.println(dp[0][n - 1]);
    }
}
class Reader
{
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader()
    {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException
    {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException
    {
        byte[] buf = new byte[64]; // line length
        int cnt = 0, c;
        while ((c = read()) != -1)
        {
            if (c == '\n')
                break;
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException
    {
        int ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do
        {
            ret = ret * 10 + c - '0';
        }  while ((c = read()) >= '0' && c <= '9');

        if (neg)
            return -ret;
        return ret;
    }

    public long nextLong() throws IOException
    {
        long ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');
        if (neg)
            return -ret;
        return ret;
    }

    public double nextDouble() throws IOException
    {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();

        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');

        if (c == '.')
        {
            while ((c = read()) >= '0' && c <= '9')
            {
                ret += (c - '0') / (div *= 10);
            }
        }

        if (neg)
            return -ret;
        return ret;
    }

    private void fillBuffer() throws IOException
    {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1)
            buffer[0] = -1;
    }

    private byte read() throws IOException
    {
        if (bufferPointer == bytesRead)
            fillBuffer();
        return buffer[bufferPointer++];
    }

    public void close() throws IOException
    {
        if (din == null)
            return;
        din.close();
    }
}