$rk$ 代表每只熊在数轴上的相对位置排名。这个拆开就是二维数点了。时间复杂度 $\mathcal{O}(n\log n)$，考验了选手处理复杂信息的能力，以及一些整体考虑，并查集和均摊的思想，还有感受结论的能力，还有把区间变成左闭右开区间的奇技淫巧，总之是个难题！

代码：

#include <bits/stdc++.h>
#define int long long
#define rep(i, l, r) for (int i (l); i <= r; ++ i)
#define rrp(i, l, r) for (int i (r); i >= l; -- i)
#define eb emplace_back
using namespace std;
#define pii pair <int, int>
#define inf 1000000000
#define ls (p << 1)
#define rs (ls | 1)
constexpr int N = 5e5 + 5, M = 1e5, P = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
inline int rd () {
  int x = 0, f = 1;
  char ch = getchar ();
  while (! isdigit (ch)) {
    if (ch == '-') f = -1;
    ch = getchar ();
  }
  while (isdigit (ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar ();
  }
  return x * f;
}
int qpow (int x, int y, int p = P) {
  int ret (1);
  for (; y; y >>= 1, x = x * x % p) if (y & 1) ret = ret * x % p;
  return ret;
}
int X[N], Y[N], n;
int xpos[N], ypos[N], xrk[N], yrk[N], lx[N], ly[N], rx[N], ry[N];
int px[N], py[N];
int dis[N];
int pre[N], nxt[N], id[N];
class DSU {
  public:
  int fa[N];
  void init () {
    rep (i, 1, n) fa[i] = i;
  }
  int find (int x) {
    return x == fa[x] ? x : fa[x] = find (fa[x]);
  }
  int merge (int x, int y) {
    if (! x || ! y) return x | y;
    x = find (x), y = find (y);
    if (x ^ y) {
      fa[x] = y; return y;
    }
    return x;
  }
} dsu, seq;
int net[N << 2], tot;
pii Mes[N << 2];
class Link {
  public: 
  int hd, tl;
  void merge (Link x) {
    if (! hd) {
      hd = x.hd, tl = x.tl;
      return ;
    }
    if (! x.hd) return ;
    net[tl] = x.hd;
    tl = x.tl;
  }
  void lnk (pii pa) {
    int x = ++ tot;
    Mes[x] = pa;
    if (! tl) {
      hd = tl = x;
      return ;
    }
    net[tl] = x;
    tl = x;
  }
} L[N], R[N];
int unite (Link x, int u) {
  for (int i (x.hd); i; i = net[i]) {
    u = dsu.merge (Mes[i].second, u);
  } return u;
}
void solve (int * x, int * p, int * rk, int * pos, int * l, int * r) {
  rep (i, 1, tot) net[i] = 0;
  dsu.init (), seq.init ();
  tot = 0;
  rep (i, 1, n) p[i] = i, l[i] = r[i] = -1;
  sort (p + 1, p + n + 1, [&] (int i, int j) { return x[i] < x[j]; });
  rep (i, 2, n) dis[i] = x[p[i]] - x[p[i - 1]]; dis[1] += n, dis[n + 1] = n + 1;
  rep (i, 1, n) rk[p[i]] = i, pre[i] = i - 1, nxt[i] = i + 1, id[i] = 0, L[i].hd = L[i].tl = 0, R[i].hd = R[i].tl = 0; nxt[0] = 1, pre[n + 1] = n;
  rep (i, 1, n) if (! dis[i]) seq.merge (i, i - 1);
  rep (T, 1, n) {
    int p = rk[T], s = seq.find (p), i; id[0] = p;
    for (i = s; i == s || ! dis[i]; i = nxt[i]) {
      nxt[s] = nxt[i], pre[nxt[i]] = s;
      if (! id[i]) continue;
      if (p >= nxt[i]) {
        L[0].merge (R[i]);
        L[0].lnk (pii (i, unite (L[i], id[i])));
      } else if (p < i) {
        R[0].merge (L[i]);
        R[0].lnk (pii (nxt[i], unite (R[i], id[i])));
      } else {
        dsu.merge (id[i], id[0]);
        int u (0);
        for (int j (L[i].hd); j; j = net[j]) {
          if (p >= Mes[j].first) u = dsu.merge (Mes[j].second, u);
          else id[0] = dsu.merge (Mes[j].second, id[0]);
        }
        L[0].lnk (pii (i, u));
        u = 0;
        for (int j (R[i].hd); j; j = net[j]) {
          if (p < Mes[j].first) u = dsu.merge (Mes[j].second, u);
          else id[0] = dsu.merge (Mes[j].second, id[0]);
        }
        R[0].lnk (pii (nxt[i], u));
      }
    }
    L[s] = L[0], R[s] = R[0], id[s] = id[0];
    L[0].hd = L[0].tl = R[0].hd = R[0].tl = id[0] = 0;
    if (! -- dis[s]) seq.merge (s, pre[s]);
    if (! -- dis[nxt[s]]) seq.merge (nxt[s], s);
    if (s)
    ++ pos[1],
    -- pos[s];
    -- pos[nxt[s]];
  }
  rep (i, 1, n) pos[i] += pos[i - 1];
  rep (i, 1, n) pos[i] += x[p[i]];
  for (int i (1); i <= n; i = nxt[i]) {
    l[id[i]] = i, r[id[i]] = nxt[i];
    for (int j (L[i].hd); j; j = net[j]) {
      l[Mes[j].second] = i, r[Mes[j].second] = Mes[j].first;
    }
    for (int j (R[i].hd); j; j = net[j]) {
      l[Mes[j].second] = Mes[j].first, r[Mes[j].second] = nxt[i];
    }
  }
  rep (i, 1, n) l[i] = l[dsu.find (i)], r[i] = r[dsu.find (i)];
}
int ans[N];
vector <pii> vec[N], vc[N];
int c[N];
int lb (int x) {
  return x & -x;
}
void upd (int x) {
  for (; x; x -= lb (x)) ++ c[x];
}
int qry (int x) {
  int ret (0);
  for (; x <= n + 1; x += lb (x)) {
    ret += c[x];
  }
  return ret;
}
int sum[N];
int32_t main () {
  // freopen ("1.in", "r", stdin);
  // freopen ("1.out", "w", stdout);
  n = rd ();
  rep (i, 1, n) X[i] = rd (), Y[i] = rd ();
  solve (X, px, xrk, xpos, lx, rx);
  solve (Y, py, yrk, ypos, ly, ry);
  int s (0);
  rep (i, 1, n) (s += 1ll * (xpos[xrk[i]] - 1) * (ypos[yrk[i]] - 1));
  rep (i, 1, n)
    vec[lx[xrk[i]]].eb (pii (ly[yrk[i]], i)),
    vec[lx[xrk[i]]].eb (pii (ry[yrk[i]], i)),
    vec[rx[xrk[i]]].eb (pii (ly[yrk[i]], i)),
    vec[rx[xrk[i]]].eb (pii (ry[yrk[i]], i));
  rrp (i, 1, n) {
    sum[i] = sum[i + 1] + xpos[xrk[py[i]]] - 1;
  }
  rep (i, 1, n) {
    ans[i] = (sum[ly[yrk[i]]] + sum[ry[yrk[i]]]);
  }
  rrp (i, 1, n) {
    sum[i] = sum[i + 1] + ypos[yrk[px[i]]] - 1;
  }
  rep (i, 1, n) {
    ans[i] = ans[i] + (sum[lx[xrk[i]]] + sum[rx[xrk[i]]]);
  }
  rrp (i, 1, n) {
    upd (yrk[px[i]]);
    for (auto p : vec[i]) {
      ans[p.second] += qry (p.first);
    }
  }
  rep (i, 1, n) {
    (ans[i] -= 1ll * (xpos[xrk[i]] - (xrk[i] < lx[xrk[i]]) + (xrk[i] >= rx[xrk[i]])) * (ypos[yrk[i]] - (yrk[i] < ly[yrk[i]]) + (yrk[i] >= ry[yrk[i]])));
    (ans[i] += 1ll * X[i] * Y[i]);
    ans[i] += s;
    printf ("%lld\n", ans[i]);
  }
}