Problem: You're given the centers of three equal sides of a strictly convex tetragon. Your task is to restore the initial tetragon.

Input

The first input line contains one number T — amount of tests (1 ≤ T ≤ 5·104). Each of the following T lines contains numbers x1, y1, x2, y2, x3, y3 — coordinates of different points that are the centers of three equal sides (non-negative integer numbers, not exceeding 10).

Output

For each test output two lines. If the required tetragon exists, output in the first line YES, in the second line — four pairs of numbers — coordinates of the polygon's vertices in clockwise or counter-clockwise order. Don't forget, please, that the tetragon should be strictly convex, i.e. no 3 of its points lie on one line. Output numbers with 9 characters after a decimal point.

If the required tetragon doen't exist, output NO in the first line, and leave the second line empty.

Examples

Input

3
1 1 2 2 3 3
0 1 1 0 2 2
9 3 7 9 9 8


Output

NO

YES
3.5 1.5 0.5 2.5 -0.5 -0.5 2.5 0.5
NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct point {
  double x, y;
};
struct line {
  double A, B, C;
};
line build(point A, point B) {
  line res;
  res.A = A.y - B.y;
  res.B = B.x - A.x;
  res.C = A.x * B.y - A.y * B.x;
  return res;
}
line per(line l, point A) {
  line res;
  res.A = -l.B;
  res.B = l.A;
  res.C = -(res.A * A.x + res.B * A.y);
  return res;
}
line par(line l, point A) {
  line res;
  res.A = l.A;
  res.B = l.B;
  res.C = -(res.A * A.x + res.B * A.y);
  return res;
}
point inter(line l1, line l2) {
  point res;
  res.x = (-l1.C * l2.B + l2.C * l1.B) / (l1.A * l2.B - l2.A * l1.B);
  res.y = (-l1.A * l2.C + l2.A * l1.C) / (l1.A * l2.B - l2.A * l1.B);
  return res;
}
point middle(point A, point B) {
  point res;
  res.x = (A.x + B.x) / 2.0;
  res.y = (A.y + B.y) / 2.0;
  return res;
}
point sym(point A, point B) {
  point res;
  res.x = 2.0 * B.x - A.x;
  res.y = 2.0 * B.y - A.y;
  return res;
}
double dist(point A, point B) {
  return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
double plos(point A, point B, point C, point D) {
  double s = fabs(A.x * B.y - B.x * A.y + B.x * C.y - B.y * C.x + C.x * D.y -
                  C.y * D.x + D.x * A.y - D.y * A.x) /
             2.0;
  return s;
}
double plos(point A, point B, point C) {
  double s = fabs(A.x * B.y - A.y * B.x + B.x * C.y - B.y * C.x + C.x * A.y -
                  C.y * A.x) /
             2.0;
  return s;
}
bool have;
void print(point A) { printf("%.9lf %.9lf", A.x, A.y); }
double Cp3(point O, point A, point B) {
  double res = (A.x - O.x) * (B.y - O.y) - (B.x - O.x) * (A.y - O.y);
  return res;
}
bool check(point A, point B, point C) { return (fabs(Cp3(A, B, C)) < 1e-9); }
bool Zero(double x) { return (fabs(x) < 1e-9 ? true : false); }
bool Dif(point A, point B, point C, point D) {
  point ans[4];
  ans[0] = A;
  ans[1] = B;
  ans[2] = C;
  ans[3] = D;
  for (int i = 0; i < 4; i++)
    if (Zero(Cp3(ans[i], ans[(i + 1) % 4], ans[(i + 2) % 4]))) return false;
  if (!Zero(plos(ans[0], ans[1], ans[2], ans[3]) -
            plos(ans[0], ans[1], ans[2]) - plos(ans[0], ans[2], ans[3])))
    return false;
  if (!Zero(plos(ans[0], ans[1], ans[2], ans[3]) -
            plos(ans[1], ans[2], ans[3]) - plos(ans[1], ans[3], ans[0])))
    return false;
  return true;
}
void solve(point A, point B, point C) {
  if (have) return;
  line l1 = per(build(A, B), middle(A, B));
  line l2 = per(build(B, C), middle(B, C));
  point D = inter(l1, l2);
  point E = sym(D, B);
  point F = inter(l1, par(l2, E));
  point G = sym(F, B);
  point H = sym(G, C);
  point I = sym(F, A);
  if (Dif(F, G, H, I) == true) {
    have = true;
    printf("YES\n");
    print(I);
    printf(" ");
    print(F);
    printf(" ");
    print(G);
    printf(" ");
    print(H);
    printf("\n");
  }
}
void solve() {
  point A, B, C;
  scanf("%lf%lf%lf%lf%lf%lf", &A.x, &A.y, &B.x, &B.y, &C.x, &C.y);
  if (check(A, B, C)) {
    printf("NO\n\n");
    return;
  }
  have = false;
  solve(A, B, C);
  solve(A, C, B);
  solve(B, C, A);
  solve(B, A, C);
  solve(C, A, B);
  solve(C, B, A);
  if (!have) {
    printf("NO\n\n");
    return;
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 1; i <= t; i++) solve();
  return 0;
}