You are given n points on a plane. 

Please find the minimum sum of areas of two axis-aligned rectangles, such that each point is contained in at least one of these rectangles.

Note that the chosen rectangles can be degenerate. Rectangle contains all the points that lie inside it or on its boundary.

Input

The first line contains one integer t (1 ≤ t ≤ 2 ⋅ 10^5) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.

The following n lines contain the coordinates of the points x_i and y_i (0 ≤ x_i, y_i ≤ 10^9). It is guaranteed that the points are distinct.

It is guaranteed that the sum of values n over all test cases does not exceed 2⋅10^5.

Output

For each test case print one integer — the minimum sum of areas.

Example

Input


3
2
9 1
1 6
2
8 10
0 7
4
0 0
1 1
9 9
10 10


Output


0
0
2

Note

In the first two test cases the answer consists of 2 degenerate rectangles. In the third test case one of the possible answers consists of two rectangles 1 × 1 with bottom left corners (0,0) and (9,9).
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
template<typename T> inline void down(T &x, T y) {x > y ? x = y : 0;}
template<typename T> inline void up(T &x, T y) {x < y ? x = y : 0;}

const int N = 2e5 + 5, K = 18;
const int inf = 1e9 + 5; 
const ll INF = 4e18;
int n, lg[N], lb, rb, db, ub;
ll ans;

struct pt {
  int x, y;
  pt(int _x = 0, int _y = 0) : x(_x), y(_y) {}
  inline void set() {x = y = -1;}
} p[N];
inline bool cmpx(const pt &x, const pt &y) {
  if (x.x != y.x) return x.x < y.x;
  return x.y < y.y;
}
inline bool cmpy(const pt &x, const pt &y) {
  if (x.y != y.y) return x.y < y.y;
  return x.x < y.x;
}

struct line {
  int l, r;
  line(int _l = 0, int _r = 0) : l(_l), r(_r) {}
  inline void set() {l = inf, r = -inf;}
  inline line operator | (const line &rhs) const {
    return line(min(l, rhs.l), max(r, rhs.r));
  }
  inline int len() {return r - l;}
} pre[N], suf[N];

inline void solve1() { // Non-intersecting -- swap x and y
  pre[0].set(), suf[n + 1].set();
  for (int i = 1; i <= n; ++i) {
    pre[i] = pre[i - 1] | line(p[i].y, p[i].y);
  }
  for (int i = n; i > 1; --i) {
    suf[i] = suf[i + 1] | line(p[i].y, p[i].y);
    down(ans, 1ll * (p[i - 1].x - p[1].x) * (pre[i - 1].r - pre[i - 1].l) + 1ll * (p[n].x - p[i].x) * (suf[i].r - suf[i].l));
  }
}

ll mn[N];
struct ST {
  ll st[K][N];
  inline void init(int s) {
    for (int k = 1; k < K; ++k)
      for (int i = s; i + (1 << k) - 1 < n; ++i)
        st[k][i] = min(st[k - 1][i], st[k - 1][i + (1 << k - 1)]);
  }
  inline ll query(int l, int r) {
    if (l > r) return INF; int t = lg[r - l + 1];
    return min(st[t][l], st[t][r - (1 << t) + 1]);
  }
} st1, st2;
inline void solve2() { // cross -- once is alright
  int k1 = rb - lb, k2 = ub - db;
  int u, d; tie(d, u) = minmax(p[1].y, p[n].y);
  sort(p + 1, p + n + 1, cmpy);
  int s1 = 0, s2 = n + 1;
  for (int i = 1; i <= n; ++i) {
    if (p[i].y <= d) up(s1, i);
    if (p[i].y >= u) down(s2, i);
  } 
  suf[n + 1].set();
//  printf("s1 = %d, s2 = %d\n", s1, s2);
  for (int i = n; i >= s2; --i) {
    suf[i] = suf[i + 1] | line(p[i].x, p[i].x);
//    printf("i = %d, l = %d, r = %d\n", i, suf[i].l, suf[i].r);
    mn[i] = 1ll * k1 * p[i].y + 1ll * k2 * (i == n ? 0 : suf[i + 1].len());
    st1.st[0][i] = 1ll * k1 * p[i].y - 1ll * k2 * suf[i + 1].l;
    st2.st[0][i] = 1ll * k1 * p[i].y + 1ll * k2 * suf[i + 1].r;
  }
  st1.init(s2), st2.init(s2), mn[s2 - 1] = INF; 
  for (int i = s2 + 1; i <= n; ++i) down(mn[i], mn[i - 1]);
  line c; c.set();
  for (int i = 1; i <= s1; ++i) {
    int pl = s2 - 1, pr = s2 - 1, cl = c.l, cr = c.r;
    for (int l = s2, r = n - 1; l <= r; ) {
      int mid = l + r >> 1; 
      if (suf[mid + 1].l <= cl) pl = mid, l = mid + 1;
      else r = mid - 1;
    } // rightmost f(i) <= c
    for (int l = s2, r = n - 1; l <= r; ) {
      int mid = l + r >> 1;
      if (suf[mid + 1].r >= cr) pr = mid, l = mid + 1;
      else r = mid - 1;
    }
//    printf("pl = %d, pr = %d\n", pl, pr);
    ll val = mn[min(pl, pr)];
    down(val, min(st1.query(pr + 1, pl) + 1ll * k2 * cr, st2.query(pl + 1, pr) - 1ll * k2 * cl)); // k1 * y_i + k2 * (cr - l_i), if pr < p 
    down(val, 1ll * k1 * p[max(pl, pr) + 1].y + 1ll * k2 * (i == 1 ? 0 : c.len()));
    down(ans, val - 1ll * k1 * p[i].y); // k1 * (down - up)
//    printf("i = %d, ans = %lld\n", i, ans);
//    printf("i = %d, val = %lld\n", i, val - 1ll * k1 * p[i].y);
    c = c | line(p[i].x, p[i].x);
  }
}

int px[N], py[N], qx[N], qy[N], c1, c2;
int pool[N << 5], *P = pool;

inline bool judge(int x1, int x2, int x3) {
  return 1ll * (py[x2] - py[x1]) * (px[x3] - px[x2]) < 1ll * (py[x3] - py[x2]) * (px[x2] - px[x1]);
} 

struct node {
  int *q, l, r;
  inline void set(int *s) {q = s, l = 0, r = -1;}
  inline void push(int i) {
    while (l < r && !judge(q[r - 1], q[r], i)) --r;
    q[++r] = i;    
  }
  inline ll query(int i) {
    while (l < r && 1ll * (py[q[l + 1]] - py[q[l]]) * qx[i] < -1ll * qy[i] * (px[q[l + 1]] - px[q[l]])) ++l;
    return 1ll * qy[i] * px[q[l]] + 1ll * qx[i] * py[q[l]];
  }
} t[N << 2];

void build(int p, int l, int r) {
  t[p].set(P), P += r - l + 1;
  for (int i = l; i <= r; ++i) t[p].push(i);
  if (l == r) return;
  int mid = l + r >> 1;
  build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r);
}

void query(int p, int l, int r, int ll, int rr, int i) {
  if (l >= ll && r <= rr) return down(ans, t[p].query(i));
  int mid = l + r >> 1;
  if (mid >= ll) query(p << 1, l, mid, ll, rr, i);
  if (mid < rr) query(p << 1 | 1, mid + 1, r, ll, rr, i);
}

struct qry {
  int l, r, i;
  inline bool operator < (const qry &q) const {
    return 1ll * qy[i] * qx[q.i] > 1ll * qy[q.i] * qx[i];
  }
} q[N];

inline void solve3() { // ??? Whatever it is -- reverse on one axis -- must intersect
  sort(p + 1, p + n + 1, cmpx), c1 = c2 = 0;
  pre[0].set(), suf[n + 1].set();
  for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] | line(p[i].y, p[i].y);
  for (int i = n; i > 1; --i) suf[i] = suf[i + 1] | line(p[i].y, p[i].y);
  for (int i = 1, last = -1; i <= n; ++i) 
    if (pre[i].l != pre[i - 1].l) {
      if (~last) qy[++c1] = last, qx[c1] = p[i].x;
      last = p[i].y;
    } // adjacent pairs
  for (int i = n, last = -1; i; --i)
    if (suf[i].r != suf[i + 1].r) {
      if (~last) px[++c2] = p[i].x, py[c2] = last;
      last = p[i].y;
    }
  if (!c1 || !c2) return;
  reverse(px + 1, px + c2 + 1), reverse(py + 1, py + c2 + 1);
  for (int i = 1; i <= c1; ++i) {
    int xb = qx[i], yb = qy[i]; // x >= xb, y >= yb
    int lp = c2 + 1, rp = 0;
    for (int l = 1, r = c2; l <= r; ) {
      int mid = l + r >> 1;
      if (py[mid] >= yb) rp = mid, l = mid + 1;
      else r = mid - 1;
    }
    for (int l = 1, r = c2; l <= r; ) {
      int mid = l + r >> 1;
      if (px[mid] >= xb) lp = mid, r = mid - 1;
      else l = mid + 1;
    }
    q[i] = (qry) {lp, rp, i};
  }
  for (int i = 1; i <= c1; ++i) qy[i] = ub - qy[i], qx[i] = rb - qx[i];
  for (int i = 1; i <= c2; ++i) px[i] -= lb, py[i] -= db;
  P = pool, build(1, 1, c2), sort(q + 1, q + c1 + 1);
  for (int i = 1; i <= c1; ++i) 
    if (q[i].l <= q[i].r) query(1, 1, c2, q[i].l, q[i].r, q[i].i);
}

inline void getbound() {
  ub = rb = -inf, lb = db = inf;
  for (int i = 1; i <= n; ++i) {
    up(ub, p[i].y), down(db, p[i].y);
    up(rb, p[i].x), down(lb, p[i].x);
  }
}

inline void solve() {
  scanf("%d", &n), ans = INF;
  for (int i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
  sort(p + 1, p + n + 1, cmpx);
  getbound(), solve1(), solve2();
  for (int i = 1; i <= n; ++i) swap(p[i].x, p[i].y);  
  sort(p + 1, p + n + 1, cmpx);
  getbound(), solve1(), solve3();
  for (int i = 1; i <= n; ++i) p[i].y *= -1;
  getbound(), solve3();
  printf("%lld\n", ans);
}

int main() {
//  freopen("in.txt", "r", stdin);
//  freopen("res.out", "w", stdout);
  for (int i = 2; i < N; ++i) lg[i] = lg[i >> 1] + 1;
  int tt; scanf("%d", &tt); while(tt--) solve();
  return 0;
}