func_code_string: #include <bits/stdc++.h>   using namespace std;   using namespace std::chrono;   template <typename A>   ostream &operator<<(ostream &cout, vector<A> const &v);   template <typename A, typename B>   ostream &operator<<(ostream &cout, pair<A, B> const &p) {     return cout <<  (  << p.first <<  ,   << p.second <<  ) ;   }   template <typename A>   ostream &operator<<(ostream &cout, vector<A> const &v) {     cout <<  [ ;     for (int i = 0; i < v.size(); i++) {       if (i) cout <<  ,  ;       cout << v[i];     }     return cout <<  ] ;   }   template <typename A, typename B>   istream &operator>>(istream &cin, pair<A, B> &p) {     cin >> p.first;     return cin >> p.second;   }   mt19937 rng(61378913);   void usaco(string filename) {     freopen((filename +  .in ).c_str(),  r , stdin);     freopen((filename +  .out ).c_str(),  w , stdout);   }   const long double pi = 3.14159265358979323846;   const long long mod = 1000000007;   namespace modop {   long long madd(long long a, long long b) { return (a + b) % mod; }   long long msub(long long a, long long b) {     return (((a - b) % mod) + mod) % mod;   }   long long mmul(long long a, long long b) {     return ((a % mod) * (b % mod)) % mod;   }   long long mpow(long long base, long long exp) {     long long res = 1;     while (exp) {       if (exp % 2 == 1) {         res = (res * base) % mod;       }       exp >>= 1;       base = (base * base) % mod;     }     return res;   }   long long minv(long long base) { return mpow(base, mod - 2); }   long long mdiv(long long a, long long b) { return mmul(a, minv(b)); }   }  // namespace modop   using namespace modop;   long long n, m, k, q, l, r, x, y, z;   long long a[1000005];   long long b[1000005];   long long c[1000005];   string second, t;   long long ans = 0;   map<long long, long long> cnt;   pair<long long, long long> vals[1000005];   void solve(int tc) {     cin >> n >> k;     cnt.clear();     for (int ele = 0; ele < n; ele++) cin >> a[ele];     ;     if (k == 1) {       cout << (n & 1) <<   n ;       return;     }     for (long long i = 0; i < (n); ++i) ++cnt[a[i]];     long long second = cnt.size();     int pos = 0;     for (pair<long long, long long> x : cnt) vals[pos++] = x;     reverse(vals, vals + second);     long long req = 0, last = vals[0].first, ans = 0;     bool fin = 0;     for (long long i = 0; i < (second); ++i) {       long long diff = last - vals[i].first;       last = vals[i].first;       for (long long j = 0; j < (diff); ++j) {         if (req == 0) break;         if (req > 1000000) {           fin = 1;           break;         }         req *= k;         if (req > 1000000) {           fin = 1;           break;         }       }       long long c = vals[i].second;       if (fin) {         ans = msub(ans, c * mpow(k, vals[i].first));         continue;       }       if (c <= req) {         req -= c;         ans = msub(ans, c * mpow(k, vals[i].first));       } else {         c -= req;         ans = 0;         req = 0;         if (c & 1) {           req = 1;           ans = (ans + mpow(k, vals[i].first)) % mod;         }       }     }     cout << (ans % mod) <<   n ;   }   int main() {     { ios_base::sync_with_stdio(false); }     {       cin.tie(NULL);       cout.tie(NULL);     }     int tc = 1;     cin >> tc;     for (int t = 0; t < tc; t++) solve(t);   }   