Codeforces Mike and palindrome
#include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(0)
#define bfast cin.tie(0)
#define outs(x) cout << x << " "
#define outn(x) cout << x << "\n"
#define sf scanf
#define pf printf
#define pfn(x , k) printf(k , x)
#define nl puts("")
#define psb push_back
#define mset(c,v) memset(c , v , sizeof c)
#define loop0(n) for(int i=0; i<n; i++)
#define loop1(n) for(int i=1; i<=n; i++)
#define mpair(x , y) make_pair(x , y)
#define all(x) x.begin(), x.end()
#define pi acos(-1.0)
#define psb push_back
#define clr clear()
typedef unsigned long long ull;
typedef long long LL;
typedef vector<int>vii;
typedef vector<LL>vll;
typedef vector<string>vs;
typedef map<int, int>mpii;
typedef map<string, int>mpsi;
typedef map<char, int>mpci;
typedef map<LL, LL>mpll;
const int mod = 1000007;
const int high = 1002;
int main()
{
fast;
string x , y;
while(cin >> x)
{
y = x;
int cnt = 0 , len = y.length();
for(int i=0, j=len-1; i<len/2; i++, j--)
{
if(x[i] != y[j]) cnt+=1;
}
if(cnt == 1 || (len%2==1 && cnt==0)) outn("YES");
else outn("NO");
}
return 0;
}
using namespace std;
#define fast ios_base::sync_with_stdio(0)
#define bfast cin.tie(0)
#define outs(x) cout << x << " "
#define outn(x) cout << x << "\n"
#define sf scanf
#define pf printf
#define pfn(x , k) printf(k , x)
#define nl puts("")
#define psb push_back
#define mset(c,v) memset(c , v , sizeof c)
#define loop0(n) for(int i=0; i<n; i++)
#define loop1(n) for(int i=1; i<=n; i++)
#define mpair(x , y) make_pair(x , y)
#define all(x) x.begin(), x.end()
#define pi acos(-1.0)
#define psb push_back
#define clr clear()
typedef unsigned long long ull;
typedef long long LL;
typedef vector<int>vii;
typedef vector<LL>vll;
typedef vector<string>vs;
typedef map<int, int>mpii;
typedef map<string, int>mpsi;
typedef map<char, int>mpci;
typedef map<LL, LL>mpll;
const int mod = 1000007;
const int high = 1002;
int main()
{
fast;
string x , y;
while(cin >> x)
{
y = x;
int cnt = 0 , len = y.length();
for(int i=0, j=len-1; i<len/2; i++, j--)
{
if(x[i] != y[j]) cnt+=1;
}
if(cnt == 1 || (len%2==1 && cnt==0)) outn("YES");
else outn("NO");
}
return 0;
}
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