Problem: A string is called diverse if it contains consecutive (adjacent) letters of the Latin alphabet and each letter occurs exactly once. For example, the following strings are diverse: "fced", "xyz", "r" and "dabcef". The following string are not diverse: "az", "aa", "bad" and "babc". Note that the letters 'a' and 'z' are not adjacent.

Formally, consider positions of all letters in the string in the alphabet. These positions should form contiguous segment, i.e. they should come one by one without any gaps. And all letters in the string should be distinct (duplicates are not allowed).

You are given a sequence of strings. For each string, if it is diverse, print "Yes". Otherwise, print "No".

Input

The first line contains integer n (1 ≤ n ≤ 100), denoting the number of strings to process. The following n lines contains strings, one string per line. Each string contains only lowercase Latin letters, its length is between 1 and 100, inclusive.

Output

Print n lines, one line per a string in the input. The line should contain "Yes" if the corresponding string is diverse and "No" if the corresponding string is not diverse. You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.

Example

Input


8
fced
xyz
r
dabcef
az
aa
bad
babc


Output


Yes
Yes
Yes
Yes
No
No
No
No

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class divs {
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			String s = sc.next();
			char[] a = s.toCharArray();
			Arrays.sort(a);
//			System.out.println(Arrays.toString(a));
			boolean flag = false;

			int sum = 0;
			if (s.length() == 1) {
				System.out.println("YES");
			} else {

				for (int i = 1; i < a.length; i++) {
					if (a[i] - a[i - 1] > 1 || (a[i] - a[i - 1] == 0)) {
						flag = false;
						break;
					} else {
						flag = true;

					}
				}

				if (flag) {
					System.out.println("YES");
				} else {
					System.out.println("NO");
				}
			}
		}
		pw.flush();
		pw.close();
	}

	/**
	 *
	 * Fixed Window Find the max sum subarray of a fixed size K
	 *
	 * Example input: [4, 2, 1, 7, 8, 1, 2, 8, 1, 0]
	 *
	 */
	public static int findMaxSumSubarray(int[] arr, int k) {
		int maxValue = Integer.MIN_VALUE;
		int currentRunningSum = 0;
		for (int i = 0; i < arr.length; i++) {
			currentRunningSum += arr[i];
			if (i >= k - 1) {
				maxValue = Math.max(maxValue, currentRunningSum);
				currentRunningSum -= arr[i - (k - 1)]; // Akher element fl 3 bto3 el K
			}
		}
		return maxValue;
	}

	// Dynamic Window
	public static int smallestSubarray(int targetSum, int[] arr) {
		int minWindowSize = Integer.MAX_VALUE;
		int currentWindowSum = 0;
		int windowStart = 0;
		for (int windowEnd = 0; windowEnd < arr.length; windowEnd++) {
			currentWindowSum += arr[windowEnd];
			while (currentWindowSum >= targetSum) {
				minWindowSize = Math.min(minWindowSize, windowEnd - windowStart + 1);
				currentWindowSum -= arr[windowStart];
				windowStart++;
			}
		}
		return minWindowSize;
	}

	// better solution for less steps
	// public static int dfs(int node) {
	// visited[node] = true;
	// int c = 1;
	// for (int i : adjList[node]) {
	// if (!visited[i]) {
	// c += dfs(i);
	// }
	// }
	// return c;
	// Getting last occur of element
	// public static int binarySearch(int low, int n, int high) {
	// low = 0;
	// high = n - 1;
	// while (low <= high) {
	// int mid = (low + high) / 2;
	// if (x >= a[mid]) {
	// if (x == a[mid]) {
	// ans = mid;
	// }
	// low = mid + 1;
	//
	// } else if (a[mid] > x) {
	// high = mid - 1;
	// }
	// }
	//
	// }
	public static int gcdFast(int x, int y) {
		if (y == 0) {
			return x;
		} else {
			return gcdFast(y, x % y);
		}
	}

	public static class Pair implements Comparable<Pair> {
		int a, b;

		public Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair A) {
			return this.a > A.a ? 1 : -1; // sorting ascending
		}
	}

	public static int findLength(String str, int k) {
		int windowStart = 0, maxLength = -1;
		HashMap<Character, Integer> hs = new HashMap<Character, Integer>();
		for (int windowEnd = 0; windowEnd < str.length(); windowEnd++) {
			char rightChar = str.charAt(windowEnd);
			hs.put(rightChar, hs.getOrDefault(rightChar, 0) + 1); // Increasing its value by 1 and if it have no value
																	// just its zero and add 1
			while (hs.size() > k) {
				char leftChar = str.charAt(windowStart);
				hs.put(leftChar, hs.get(leftChar) - 1);
				if (hs.get(leftChar) == 0) {
					hs.remove(leftChar);
				}
				windowStart++;
			}
			maxLength = Math.max(maxLength, windowEnd - windowStart + 1);
		}
		return maxLength;
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(String file) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(file));
		}

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

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

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

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

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

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

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