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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void print(T x) {
  cout << x << endl;
}
template <typename T1, typename T2>
void print2(T1 x, T2 y) {
  cout << x << " " << y << endl;
}
template <typename T1, typename T2, typename T3>
void print3(T1 x, T2 y, T3 z) {
  cout << x << " " << y << " " << z << endl;
}
bool prime[100050 + 1];
void sieve1() {
  for (int i = 0; i <= 100050; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (long long int i = 4; i <= 100050; i += 2) prime[i] = 0;
  for (int p = 3; p * p <= 100050; p += 2) {
    if (prime[p] == 1)
      for (int i = p * 2; i <= 100050; i += p) {
        prime[i] = 0;
      }
  }
}
bool isvowel(char v) { return (0x208222 >> (v & 0x1f)) & 1; }
long long int mceil(long long int a, long long int b) {
  if (a % b == 0)
    return (a / b);
  else
    return (a / b + 1);
}
long long int mfloor(long long int a, long long int b) { return (a / b); }
long long modmul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long modadd(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007) + 1000000007) % 1000000007;
}
long long modsub(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long int fastexpo(long long int a, long long int b) {
  a = a % 1000000007;
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1ll * a) % 1000000007;
    a = (a * 1ll * a) % 1000000007;
    b = b / 2;
  }
  return ans;
}
long long int cntbits(long long int n) {
  long long int cnt = 0;
  while (n) {
    if (n % 2) cnt++;
    n /= 2;
  }
  return cnt;
}
void chandan1() {
  int y = 1;
  return;
}
void chandan2() {
  for (long long int i = 0; i < 10; i++) {
    long long int x = 1;
  }
  return (chandan1());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  chandan2();
  while (t--) {
    long long int n;
    string s;
    cin >> s;
    sort(s.begin(), s.end());
    long long int f = 0;
    for (long long int i = 1; i < int(s.size()); i++) {
      if (s[i] != s[i - 1] + 1) {
        cout << "No\n";
        f = 1;
        break;
      }
    }
    if (f == 0) cout << "Yes\n";
  }
  return 0;
}