Problem: A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.

You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of characters in the string.

The second line contains the string. The string consists only of uppercase and lowercase Latin letters.

Output

Output "YES", if the string is a pangram and "NO" otherwise.

Examples

Input

12
toosmallword


Output

NO


Input

35
TheQuickBrownFoxJumpsOverTheLazyDog


Output

YES

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class A {
	public static void main(String[] args) {
		FastScanner fs=new FastScanner();
		fs.next();
		String alpha="abcdefghijklmnopqrstuvwxyz";
		String alphaUpper=alpha.toUpperCase();
		boolean[] hits=new boolean[26];
		String line=fs.next();
		for (char c:line.toCharArray())
			if (alpha.contains(c+""))
				hits[alpha.indexOf(c+"")]=true;
			else
				hits[alphaUpper.indexOf(c+"")]=true;
		for (boolean b:hits) {
			if (!b) {
				System.out.println("NO");
				return;
			}
		}
		System.out.println("YES");
	}
	
	private static boolean possible(int n, int rocketWeight, int[] planetLiftoffWeightPerFuel, int[] planetLandWeightPerFuel, double fuel) {
		for (int i=0; i<n; i++) {
			double newCost=(rocketWeight+fuel)/planetLiftoffWeightPerFuel[i];
			if (newCost>fuel) return false;
			fuel-=newCost;
			newCost=(rocketWeight+fuel)/planetLandWeightPerFuel[(i+1)%n];
			if (newCost>fuel) return false;
			fuel-=newCost;
		}
		return true;
	}
	
	private static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		public FastScanner() {
			br = new BufferedReader(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 nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++)
				a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++)
				a[i]=nextLong();
			return a;
		}
	}
}