#include #define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i)) #define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i)) #define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i)) #define REP3R(i, m, n) for (int i = int(n) - 1; (i) >= (int)(m); -- (i)) #define ALL(x) begin(x), end(x) #define dump(x) cerr << #x " = " << x << endl using ll = long long; using namespace std; template struct mint { int64_t value; // faster than int32_t a little mint() = default; // value is not initialized mint(int64_t value_) : value(value_) {} // assume value is in proper range inline mint operator + (mint other) const { int64_t c = this->value + other.value; return mint(c >= MOD ? c - MOD : c); } inline mint operator - (mint other) const { int64_t c = this->value - other.value; return mint(c < 0 ? c + MOD : c); } inline mint operator * (mint other) const { int64_t c = this->value * int64_t(other.value) % MOD; return mint(c < 0 ? c + MOD : c); } inline mint & operator += (mint other) { this->value += other.value; if (this->value >= MOD) this->value -= MOD; return *this; } inline mint & operator -= (mint other) { this->value -= other.value; if (this->value < 0) this->value += MOD; return *this; } inline mint & operator *= (mint other) { this->value = this->value * int64_t(other.value) % MOD; if (this->value < 0) this->value += MOD; return *this; } inline mint operator - () const { return mint(this->value ? MOD - this->value : 0); } mint pow(uint64_t k) const { mint x = *this, y = 1; for (; k; k >>= 1) { if (k & 1) y *= x; x *= x; } return y; } mint inv() const { return pow(MOD - 2); } // MOD must be a prime inline mint operator / (mint other) const { return *this * other.inv(); } inline mint operator /= (mint other) { return *this *= other.inv(); } inline bool operator == (mint other) const { return value == other.value; } inline bool operator != (mint other) const { return value != other.value; } }; constexpr int MOD = 1e9 + 7; mint solve(int n, ll k, vector a) { sort(ALL(a)); mint acc = 1; int l = 0; int r = n - 1; REP (i, n / 2) { int r = n - i - 1; while (l < r and a[l] + a[r] <= k) ++ l; if (l >= r) { acc *= n - 2 * i - 1; } else { acc *= l - i; } } return acc; } int main() { int n; ll k; cin >> n >> k; vector a(n); REP (i, n) cin >> a[i]; cout << solve(n, k, a).value << endl; return 0; }