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:
//package ;
import java.io.*;
import java.util.*;

public class D {
	static int n,dp[][];
	static char[]c;
	static int dp(int l,int r)
	{
		if( l > r ) return 0;
		
		if(dp[l][r]!=-1) return dp[l][r];
		
		int ans=dp(l+1,r)+1;
		for(int i=l+1;i<=r;i++)
		{
			if(c[l]==c[i])
				ans=Math.min(ans, dp(l+1,i-1)+dp(i,r));
		}
		return dp[l][r]=ans;
	}
	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter pw = new PrintWriter(System.out);
		n=sc.nextInt();
		c=sc.nextCharArray();
		dp=new int[n][n];
		for(int i=0;i<n;i++)
			Arrays.fill(dp[i], -1);
		System.out.println(dp(0,n-1));
		pw.close();
	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}

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

		long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		boolean hasnext() throws IOException {
			return br.ready();
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] a = new Integer[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		char[] nextCharArray() throws IOException {
			return br.readLine().toCharArray();
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] ans = new double[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextDouble();
			return ans;
		}

	}
}