Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That's why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.

To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol a, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with b, and so on. Kostya is well-mannered, so he doesn't use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.

You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya's obfuscation.

Input

In the only line of input there is a string S of lowercase English letters (1 ≤ |S| ≤ 500) — the identifiers of a program with removed whitespace characters.

Output

If this program can be a result of Kostya's obfuscation, print "YES" (without quotes), otherwise print "NO".

Examples

Input

abacaba


Output

YES


Input

jinotega


Output

NO

Note

In the first sample case, one possible list of identifiers would be "number string number character number string number". Here how Kostya would obfuscate the program:

  * replace all occurences of number with a, the result would be "a string a character a string a",
  * replace all occurences of string with b, the result would be "a b a character a b a",
  * replace all occurences of character with c, the result would be "a b a c a b a",
  * all identifiers have been replaced, thus the obfuscation is finished.
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (long long)((1e5) + 100);
long long cuberoot(long long x) {
  long long lo = 1, hi = min(2000000ll, x);
  while (hi - lo > 1) {
    long long mid = (lo + hi) / 2;
    if (mid * mid * mid < x) {
      lo = mid;
    } else
      hi = mid;
  }
  if (hi * hi * hi <= x)
    return hi;
  else
    return lo;
}
const long long dx[4] = {-1, 1, 0, 0};
const long long dy[4] = {0, 0, -1, 1};
long long XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const long long N = (long long)(1 * 1e6 + 10);
bool comp(pair<long long, pair<long long, long long>> p1,
          pair<long long, pair<long long, long long>> p2) {
  if (p1.first > p2.first)
    return true;
  else if (p1.second == p2.second) {
    if (p1.second.first < p2.second.first) return true;
  }
  return false;
}
long long fact(long long n);
long long ncr(long long n, long long r) {
  return fact(n) / (fact(r) * fact(n - r));
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = res * i;
  return res % 1000000007;
}
const long long a = 1000000000;
long long nCr(long long n, long long r) {
  long long fac1 = 1, fac2 = 1, fac;
  for (long long i = r; i >= 1; i--, n--) {
    fac1 = fac1 * n;
    if (fac1 % i == 0)
      fac1 = fac1 / i;
    else
      fac2 = fac2 * i;
  }
  fac = fac1 / fac2;
  return fac % a;
}
long long knapsack(long long val[], long long wt[], long long W, long long n) {
  if (n == 0 || W == 0) return 0;
  if (wt[n - 1] > W)
    return knapsack(val, wt, W, n - 1);
  else
    return max(val[n - 1] + knapsack(val, wt, W - wt[n - 1], n - 1),
               knapsack(val, wt, W, n - 1));
}
const long long m = 1000000007;
long long tree[10000];
void build(long long node, long long tl, long long tr, long long arr[]) {
  if (tr == tl) {
    tree[node] = arr[tl];
    return;
  }
  long long mid = (tr + tl) / 2;
  build(2 * node, tl, mid, arr);
  build(2 * node + 1, mid + 1, tr, arr);
  tree[node] = tree[2 * node] + tree[2 * node + 1];
}
long long query_sum(long long node, long long tl, long long tr, long long l,
                    long long r) {
  if (l > tr || r < tl) return 0;
  if (tl >= l && tr <= r) return tree[node];
  long long mid = (tl + tr) / 2;
  long long p1 = query_sum(2 * node, tl, mid, l, r);
  long long p2 = query_sum(2 * node + 1, mid + 1, tr, l, r);
  return p1 + p2;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
signed main() {
  string s;
  char cr = 'a';
  cin >> s;
  long long cnt = 0;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == cr) cr++;
    if (s[i] > cr) {
      cout << "NO"
           << "\n";
      return 0;
    }
  }
  cout << "YES"
       << "\n";
}
