#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
namespace IO {
template <class A, class B>
ostream &operator<<(ostream &out, vector<pair<A, B>> a);
template <class A>
ostream &operator<<(ostream &out, vector<A> a);
template <class A, class B>
ostream &operator<<(ostream &out, pair<A, B> a) {
  out << a.first << " " << a.second;
  return out;
}
template <class A, class B>
ostream &operator<<(ostream &out, vector<pair<A, B>> a) {
  for (pair<A, B> x : a) out << x.first << " " << x.second << '\n';
  return out;
}
template <class A>
ostream &operator<<(ostream &out, vector<A> a) {
  for (A x : a) out << x << ' ';
  return out;
}
template <class A, class B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
template <class A>
istream &operator>>(istream &in, vector<A> &a) {
  for (A &x : a) in >> x;
  return in;
}
}  // namespace IO
using namespace IO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long x = 0;
  for (long long i = (0); i < (long long)(n); ++i) x ^= a[i];
  if (x == 0) {
    cout << ("DRAW") << '\n';
    return;
  }
  long long lat_bit = 32;
  while (!((1ll << lat_bit) & x)) --lat_bit;
  long long cnt = 0;
  for (long long i = (0); i < (long long)(n); ++i)
    cnt += ((1ll << lat_bit) & a[i]) > 0;
  long long rem = n - cnt;
  if (cnt == 1) {
    cout << ("WIN") << '\n';
    return;
  }
  if ((cnt / 2) % 2) {
    cout << (rem % 2 ? "WIN" : "LOSE") << '\n';
  } else {
    cout << ("WIN") << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(30);
  long long q = 1;
  cin >> q;
  while (q--) solve();
}
