#pragma GCC optimize("O3") //コンパイラ最適化用

//#define _GLIBCXX_DEBUG //配列に[]でアクセス時のエラー表示
/* #region   */
#define _USE_MATH_DEFINES
#include <algorithm> //sort,二分探索,など
#include <bitset>    //固定長bit集合
// #include <boost/multiprecision/cpp_dec_float.hpp>
// #include <boost/multiprecision/cpp_int.hpp>
#include <cassert> //assert(p)で，not pのときにエラー
#include <cctype>
#include <chrono> //実行時間計測
#include <climits>
#include <cmath>   //pow,logなど
#include <complex> //複素数
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional> //sortのgreater
#include <iomanip>    //setprecision(浮動小数点の出力の誤差)
#include <ios>        // std::left, std::right
#include <iostream>   //入出力
#include <iterator>   //集合演算(積集合,和集合,差集合など)
#include <map>
#include <numeric> //iota(整数列の生成),gcdとlcm,accumulate
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility> //pair
#include <vector>
using namespace std;
typedef long long LL;
typedef long double LD;

#define ALL(x) x.begin(), x.end()
const long long INF = numeric_limits<long long>::max() / 4;
const int MOD = 1e9 + 7;
// const int MOD=998244353;
//略記
#define FF first
#define SS second
#define int long long
#define stoi stoll
#define LD long double
#define PII pair<int, int>
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(x) (int)((x).size())
#define VB vector<bool>
#define VVB vector<vector<bool>>
#define VI vector<int>
#define VVI vector<vector<int>>

#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REPD(i, n) for (int i = (int)(n) - (int)1; i >= 0; i--)
#define FOR(i, a, b) for (int i = a; i < (int)(b); i++)
#define FORD(i, a, b) for (int i = (int)(b) - (int)1; i >= (int)a; i--)

const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1},
          Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};

int in() {
    int x;
    cin >> x;
    return x;
}
// https://qiita.com/Lily0727K/items/06cb1d6da8a436369eed#c%E3%81%A7%E3%81%AE%E5%AE%9F%E8%A3%85
void print() { cout << "\n"; }

template <class Head, class... Tail> void print(Head &&head, Tail &&...tail) {
    cout << head;
    if (sizeof...(tail) != 0)
        cout << " ";
    print(forward<Tail>(tail)...);
}

template <class T> void print(vector<T> &vec) {
    for (auto &a : vec) {
        cout << a;
        if (&a != &vec.back())
            cout << " ";
    }
    cout << "\n";
}

template <class T> void print(set<T> &set) {
    for (auto &a : set) {
        cout << a << " ";
    }
    cout << "\n";
}

template <class T> void print(vector<vector<T>> &df) {
    for (auto &vec : df) {
        print(vec);
    }
}
long long power(long long x, long long n) {
    // O(logn)
    // https://algo-logic.info/calc-pow/#toc_id_1_2
    long long ret = 1;
    while (n > 0) {
        if (n & 1)
            ret *= x; // n の最下位bitが 1 ならば x^(2^i) をかける
        x *= x;
        n >>= 1; // n を1bit 左にずらす
    }
    return ret;
}
// @brief nCr. O(r log n)。あるいは前処理 O(n), 本処理 O(1)で求められる modint
// の bc を検討。
int comb(int n, int r) {
    // https://www.geeksforgeeks.org/program-to-calculate-the-value-of-ncr-efficiently/

    // p holds the value of n*(n-1)*(n-2)...,
    // k holds the value of r*(r-1)...
    long long p = 1, k = 1;

    // C(n, r) == C(n, n-r),
    // choosing the smaller value
    if (n - r < r)
        r = n - r;

    if (r != 0) {
        while (r) {
            p *= n;
            k *= r;

            // gcd of p, k
            long long m = __gcd(p, k);

            // dividing by gcd, to simplify
            // product division by their gcd
            // saves from the overflow
            p /= m;
            k /= m;

            n--;
            r--;
        }

        // k should be simplified to 1
        // as C(n, r) is a natural number
        // (denominator should be 1 ) .
    }

    else
        p = 1;

    // if our approach is correct p = ans and k =1
    return p;
}

// MOD
void add(long long &a, long long b) {
    a += b;
    if (a >= MOD)
        a -= MOD;
}
template <class T> inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

// 負数も含む丸め
long long ceildiv(long long a, long long b) {
    if (b < 0)
        a = -a, b = -b;
    if (a >= 0)
        return (a + b - 1) / b;
    else
        return a / b;
}
long long floordiv(long long a, long long b) {
    if (b < 0)
        a = -a, b = -b;
    if (a >= 0)
        return a / b;
    else
        return (a - b + 1) / b;
}
long long floorsqrt(long long x) {
    assert(x >= 0);
    long long ok = 0;
    long long ng = 1;
    while (ng * ng <= x)
        ng <<= 1;
    while (ng - ok > 1) {
        long long mid = (ng + ok) >> 1;
        if (mid * mid <= x)
            ok = mid;
        else
            ng = mid;
    }
    return ok;
}

// @brief a^n mod mod
long long modpower(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// @brief s が c を含むか
template <class T> bool contain(const std::string &s, const T &c) {
    return s.find(c) != std::string::npos;
}

__attribute__((constructor)) void faster_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
}

/* #endregion */
int p;

int get_grundy(int n, int k) {
    VI grundy(n, -1);
    grundy[n - 1] = 0; // 負け確局面
    VI cnt(max(n, k) + 10, 0);
    set<int> got;
    FORD(ni, max(0LL, n - 1 - k), n - 1) {
        got.insert(grundy[ni + 1]);
        int mex = 0;
        while (got.count(mex)) {
            mex++;
        }
        grundy[ni] = mex;
    }
    // print(grundy);
    // exit(0);

    FORD(ni, max(0LL, n - 1 - k), n - 1) { cnt[grundy[ni]]++; }
    // print(cnt);
    // exit(0);

    REPD(ni, max(0LL, n - 1 - k)) {
        /*         if (grundy[ni + k + 1] >= k + 10) {
                    print(ni);
                    exit(0);
                }
         */
        cnt[grundy[ni + k + 1]]--;
        cnt[grundy[ni + 1]]++;

        int mex = 0;
        while (cnt[mex] >= 1) {
            mex++;
        }
        grundy[ni] = mex;
    }
    // print(grundy);
    return grundy[0];
}

void solve(int n, int k) {
    if (get_grundy(n, k) == 0) {
        print("Lose");
    } else {
        print("Win");
    }
}

signed main() {
    cin >> p;
    VI N(p), K(p);
    REP(pi, p) { cin >> N[pi] >> K[pi]; }
    REP(pi, p) { solve(N[pi], K[pi]); }
    return 0;
}