Educational Codeforces Round 39 (Rated for Div. 2) C. String Transformation

题目链接：点击打开链接

C. String Transformation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

aacceeggiikkmmooqqssuuwwyy

output

abcdefghijklmnopqrstuvwxyz

input

Copy

thereisnoanswer

output

-1

题意：给一个字符串，你可以对任意一个位置的字符，运用"膜法"，把这个字符变成阿斯克码+1的字符，任意一个位置的字符可以多次使用“膜法，”膜法“只能从小往大变。然后判断是否可以出现一个子序列为a-z。

题解：当时写的时候卡题意卡了十年。一直没明白这个”膜法“的具体效果。明白”膜法“的效果之后。先定义一个标记，标记的用途是我们现在在找a-z中那个字符。我们只需要字符串从头扫一边，边扫边判断这个位置是否能变成标记。能的话标记+1，然后继续向后找。若在遍历结束之前标记等于z了。那么就是可以，否者就是不行。

具体实现过程看代码注释：

#include
#include
#include
#include
using namespace std;

int main(){
	char s[100005]; 
	cin >> s;
	int len = strlen(s);
	int flag = 0;
	if(len <= 25){  // 若长度小于25那么一定没有a-z的子序列 
		cout << "-1" << endl;
		return 0;
	}
	int ans = 97;  // 标记 用来表示现在在找那个字符。定义为97的原因是a的阿斯克码是97 
	int a[26];  // 记录那个位置变成了那个字符，因为我们要输出 变化后的序列 
	memset(a,0,sizeof(a));
	int cur = 0 ;
	for(int i = 0 ; i < len ; i ++){  // 运用“膜法”中 
		if(cur == 26)
			break;
		if(s[i] <= ans){
			ans++;
			a[cur++] = i;
		}
		
	}
	if(a[25]){
		int num = 0;
		for(int i = 0 ; i < len ; i ++){
			if(i == a[num]){
				printf("%c",num+97);
				num++;
			}
			else 
				printf("%c",s[i]);	
		}
	}
	else 
		 cout << "-1" << endl;
}