#include #include #define rep(i,n) for(int i=0;i vi; typedef vector vl; typedef vector> vvi; typedef vector> vvl; typedef long double ld; typedef pair P; ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const static_modint& a) {os << a.val(); return os;} template ostream& operator<<(ostream& os, const dynamic_modint& a) {os << a.val(); return os;} template istream& operator>>(istream& is, vector& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template ostream& operator<<(ostream& os, const pair& p){os << p.first << ' ' << p.second; return os;} template ostream& operator<<(ostream& os, const vector& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const vector>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template ostream& operator<<(ostream& os, const set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const unordered_set& se){for(T x : se) os << x << " "; os << "\n"; return os;} template ostream& operator<<(ostream& os, const atcoder::segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template ostream& operator<<(ostream& os, const atcoder::lazy_segtree& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template void chmin(T& a, T b){a = min(a, b);} template void chmax(T& a, T b){a = max(a, b);} int solve(){ int n, m; cin >> n >> m; m--; vector ans(n); using T = tuple; priority_queue pq; auto divide = [&](int l, int r, int pos){ pq.emplace((pos - l + 1) / 2, -l, -pos); pq.emplace((r - (pos + 1) + 1) / 2, -(pos + 1), -r); }; ans[m] = 1; pq.emplace(m, -0, -m); pq.emplace((n - 1) - m, -(m + 1), -n); for(int num = 2; num <= n; num++){ auto [tmp, _l, _r] = pq.top(); pq.pop(); int l = -_l; int r = -_r; if(l == 0){ ans[l] = num; divide(l, r, l); }else if(r == n){ ans[r - 1] = num; divide(l, r, r - 1); }else{ int pos = l + (r - l - 1) / 2; ans[pos] = num; divide(l, r, pos); } // cout << tmp << ' '<< l << ' ' << r << "\n"; } cout << ans; return 0; } int main(){ int t; cin >> t; rep(i, t) solve(); return 0; }