798A 还是很水...分析情况要分析完整哦

本文介绍了一种算法,该算法检查通过更改一个字符是否可以使给定的字符串变成回文串。文章提供了两个实现示例,并解释了如何通过计数不匹配的字符来确定字符串是否能被转换。

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ike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.

A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.

Input

The first and single line contains string s (1 ≤ |s| ≤ 15).

Output

Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.

Example
Input

abccaa

Output

YES

Input

abbcca

Output

NO

Input

abcda

Output

YES

分奇偶奇偶~~~~
注意奇数时候
他中间呢个可以变的呢
奇数的回文串变一个还回文诶嘤嘤嘤

WAWAWA:::::::


!!//WAWAWA:::::::
#include <iostream>
#include <string>
using namespace std;
string s;
int main()
{

    cin >> s;
    int  cnt = 0;
    int len = s.size();
    for (int i = 0;i<len/2;i++)
        if (s[i] != s[len - i - 1])
            cnt++;
        if (cnt == 1) {
        cout << "YES";
        return 0;
    }
    cout << "NO";
    return 0;
}




//ACACAC:::::::::::::

#include <iostream>
#include <string>
using namespace std;
string s;
int main()
{

    cin >> s;
    int  cnt = 0;
    int len = s.size();
    for (int i = 0;i<len/2;i++)
        if (s[i] != s[len - i - 1])
            cnt++;
        if (cnt == 1||(s.size()%2==1 && cnt==0)) {
        cout << "YES";
        return 0;
    }
    cout << "NO";
    return 0;
}
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