第一行u表示数据组数。对于每组数据，第一行N表示石子堆数，第二行N个数ai表示第i堆石子的个数(a1<=a2<=……<=an)。 1<=u<=10 1<=n<=1000 0<=ai<=10000

Output

u行，若先手必胜输出TAK，否则输出NIE。

Sample Input2 2 2 2 3 1 2 4

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {ll x = 0;char c = getchar();bool f = false;while (!isdigit(c)) {if (c == '-') f = true;c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f ? -x : x;
}ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {if (!b) {x = 1; y = 0; return a;}ans = exgcd(b, a%b, x, y);ll t = x; x = y; y = t - a / b * y;return ans;
}
*/int T;
int n; int a[maxn];
int pp[maxn];int main() {ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin >> T;while (T--) {int ans = 0;cin >> n;for (int i = 1; i <= n; i++)cin >> a[i];for (int i = 1; i <= n; i++) {pp[i] = a[i] - a[i - 1];}for (int i = n; i >= 1; i -= 2) {ans ^= pp[i];}if (ans)cout << "TAK" << endl;else cout << "NIE" << endl;}return 0;
}