s-palindrome

s-palindrome

Let's call a string "s-palindrome" if it is symmetric about the middle of the string. For example, the string "oHo" is "s-palindrome", but the string "aa" is not. The string "aa" is not "s-palindrome", because the second half of it is not a mirror reflection of the first half.

English alphabet

You are given a string s. Check if the string is "s-palindrome".

Input

The only line contains the string s (1 ≤ |s| ≤ 1000) which consists of only English letters.

Output

Print "TAK" if the string s is "s-palindrome" and "NIE" otherwise.

Examples

input

oXoxoXo

output

TAK

input

bod

output

TAK

input

ER

output

NIE
分析：考眼力题；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=7e5+10;
const int mod=1e6+3;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,cnt,ok;
map<char,int>a;
string b;
bool check(char p,char q)
{
    if(p>q)swap(p,q);
    if(p=='b'&&q=='d')return true;
    if(p=='p'&&q=='q')return true;
    return false;
}
int main()
{
    int i,j,k,t;
    cin>>b;
    a['A']=1,a['H']=1,a['I']=1,a['M']=1,a['M']=1,a['O']=1,a['o']=1,a['T']=1,a['U']=1,a['V']=1,a['v']=1,a['W']=1,a['w']=1,a['X']=1,a['x']=1,a['Y']=1;
    int len=b.length();
    rep(i,0,len/2)if((!a[b[i]]||b[i]!=b[len-1-i])&&!check(b[i],b[len-1-i]))return 0*puts("NIE");
    puts("TAK");
    //system ("pause");
    return 0;
}