Problem: You are given a string s consisting of |s| small english letters.

In one move you can replace any character of this string to the next character in alphabetical order (a will be replaced with b, s will be replaced with t, etc.). You cannot replace letter z with any other letter.

Your target is to make some number of moves (not necessary minimal) to get string abcdefghijklmnopqrstuvwxyz (english alphabet) as a subsequence. Subsequence of the string is the string that is obtained by deleting characters at some positions. You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence or say that it is impossible.

Input

The only one line of the input consisting of the string s consisting of |s| (1 ≤ |s| ≤ 105) small english letters.

Output

If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it. Otherwise print «-1» (without quotes).

Examples

Input

aacceeggiikkmmooqqssuuwwyy


Output

abcdefghijklmnopqrstuvwxyz


Input

thereisnoanswer


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
const double eps = 1e-9;
long long INF = 1e18 + 7;
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first < b.first)
    return true;
  else if (a.first > b.first)
    return false;
  else
    return a.first < b.first;
}
struct compare {
  bool operator()(const long long& l, const long long& r) { return l > r; }
};
void read(string str, int node) {
  stringstream ss;
  int number;
  ss << str;
  while (ss >> number) cout << number << endl;
}
long long fp(long long b, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) return fp((b * b) % INF, p / 2);
  return (fp(b, p - 1) * b) % INF;
}
unsigned long long read_binary(string s) {
  unsigned long long x = 0, tmp = 1;
  for (int i = s.size() - 1; i >= 0; i--) {
    x = (x + tmp * (s[i] - '0')) % INF;
    tmp = (tmp * 2) % INF;
  }
  return x;
}
long long com(long long x, long long y) {
  long long sum = 1;
  for (int i = 1; i <= y; i++) {
    sum *= x;
    x--;
    sum /= i;
  }
  return sum;
}
long long gcd(long long n1, long long n2) {
  if (n2 != 0)
    return gcd(n2, n1 % n2);
  else
    return n1;
}
bool sum[1500];
void sieve() {
  sum[0] = sum[1] = 1;
  for (int i = 2; i * i < 1002; i++) {
    if (!sum[i]) {
      for (int j = 2 * i; j < 1002; j += i) {
        sum[j] = 1;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  string s;
  cin >> s;
  if (s.size() < 26) return cout << -1 << endl, 0;
  queue<char> q;
  for (char i = 'a'; i <= 'z'; i++) q.push(i);
  string res;
  for (int i = 0; i < s.size(); i++) {
    if (q.empty()) break;
    char x = q.front();
    if (s[i] <= q.front())
      res += q.front(), q.pop();
    else
      res += s[i];
  }
  if (q.empty()) {
    for (int i = res.size(); i < s.size(); i++) res += s[i];
    cout << res << endl;
  } else
    cout << -1 << endl;
}