Koa the Koala and her best friend want to play a game.

The game starts with an array a of length n consisting of non-negative integers. Koa and her best friend move in turns and each have initially a score equal to 0. Koa starts.

Let's describe a move in the game:

  * During his move, a player chooses any element of the array and removes it from this array, xor-ing it with the current score of the player.

More formally: if the current score of the player is x and the chosen element is y, his new score will be x ⊕ y. Here ⊕ denotes [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).

Note that after a move element y is removed from a.

  * The game ends when the array is empty. 



At the end of the game the winner is the player with the maximum score. If both players have the same score then it's a draw.

If both players play optimally find out whether Koa will win, lose or draw the game.

Input

Each test contains multiple test cases. The first line contains t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains the integer n (1 ≤ n ≤ 10^5) — the length of a.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — elements of a.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case print:

  * WIN if Koa will win the game. 
  * LOSE if Koa will lose the game. 
  * DRAW if the game ends in a draw. 

Examples

Input


3
3
1 2 2
3
2 2 3
5
0 0 0 2 2


Output


WIN
LOSE
DRAW


Input


4
5
4 1 5 1 3
4
1 0 1 6
1
0
2
5 4


Output


WIN
WIN
DRAW
WIN

Note

In testcase 1 of the first sample we have:

a = [1, 2, 2]. Here Koa chooses 1, other player has to choose 2, Koa chooses another 2. Score for Koa is 1 ⊕ 2 = 3 and score for other player is 2 so Koa wins.
// Parasparopagraho Jīvānām

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std ;
using namespace  __gnu_pbds ;
using namespace __gnu_cxx ;

typedef long long ll ;
typedef long double ld ;

typedef pair<int, int> pii ;
typedef pair<int, ll> pil ;
typedef pair<ll,ll> pll ;

typedef vector<int> vi ;
typedef vector<ll> vll ;
typedef vector<pii> vpi ;
typedef vector<pll> vpl ;
typedef vector<int,pil> edges ;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define rep(i,a,b) for (int i = (a); i <= (b); i++)
#define per(i,b,a) for (int i = (b); i >= (a); i--)

#define mp make_pair
#define eb emplace_back
#define pb push_back
#define fi first
#define se second

#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)

const int mod1 = 1000000007 ;
const int mod2 = 998244353 ;
const ll  infl = 2e18 ;
const int infi = 2e9 ;
const int maxn = 2e5 + 5 ;
const int block = 500 ;
const int logn = 31 ;
const int alpha = 27 ;
const ld pi = 3.141592653589793238462643383279 ;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()) ;

/* Careful when using long long
__builtin_clz(int x) {} //: the number of zeros at the beginning of the number
__builtin_ctz(int x) {} //: the number of zeros at the end of the number
__builtin_popcount(int x) {} //: the number of ones in the number
__builtin_parity(int x) {} //: the parity (even or odd) of the number of ones*/
ll modpow(ll a, ll b, int MOD) { ll res = 1 ; while(b) { if(b&1) { res = res*a ; res %= MOD ; } a *= a ; a %= MOD ; b >>= 1 ; } return res%MOD ; }
void upmin(int &a, int b) { if(a < b) { a = b ; } }
void relax(int &a, int b) { if(a > b) { a = b ; } }
ll add(ll a, ll b, int MOD) { a += b ; if(a >= MOD) { a -= MOD ; } return  a ; }
ll sub(ll a, ll b, int MOD) { a -= b ; if(a < 0) { a += MOD ; } return a ; }
ll mul(ll a, ll b, int MOD) { b %= MOD ; a *= b ; a %= MOD ; return a ; }
ll inverse(ll a, ll MOD) { a = modpow(a, MOD - 2, MOD) ; return a ; }
ll lcm(ll a, ll b) { ll ret ; ll g = __gcd(a, b) ; ret = a/g ; ret = ret*b ; return ret ; }


int main()
{
    ios::sync_with_stdio(false) ;
    cin.tie(NULL) ;
    int tc ;
    cin >> tc ;
    while(tc--)
    {
        int n ;
        cin >> n ;
        ll a[n] ;
        rep(i,0,n-1)
        {
            cin >> a[i] ;
        }
        int cnt[logn+5] ;
        int msb[logn+5] ;
        rep(i,0,logn)
        {
            cnt[i] = 0 ;
            msb[i] = 0 ;
        }
        rep(i,0,n-1)
        {
            if(a[i] == 0) continue ;
            int m = 0 ;
            rep(j,0,logn)
            {
                if((a[i] >> j)&1)
                {
                    cnt[j]++ ;
                    m = j ;
                }
            }
            msb[m]++ ;
        }
        int state = -1 ;
        per(j,logn,0)
        {
            if(cnt[j]%2 == 0) continue ;
            if(msb[j] == 1)
            {
                state = 1 ;
                break ;
            }
            int rem = n - cnt[j] ;
            if(rem&1)
            {
               state = 1 ;
               break ;
            }
            else
            {
                state = 2 ;
                break ;
            }
        }
        if(state == -1) cout << "DRAW\n" ;
        if(state == 1) cout << "WIN\n" ;
        if(state == 2) cout << "LOSE\n" ;
    }

    return 0 ;
}
