Problem: On the surface of a newly discovered planet, which we model by a plane, explorers found remains of two different civilizations in various locations. They would like to learn more about those civilizations and to explore the area they need to build roads between some of locations. But as always, there are some restrictions: 

  1. Every two locations of the same civilization are connected by a unique path of roads 
  2. No two locations from different civilizations may have road between them (explorers don't want to accidentally mix civilizations they are currently exploring) 
  3. Roads must be straight line segments
  4. Since intersections are expensive to build, they don't want any two roads to intersect (that is, only common point for any two roads may be at some of locations) 



Obviously all locations are different points in the plane, but explorers found out one more interesting information that may help you – no three locations lie on the same line!

Help explorers and find a solution for their problem, or report it is impossible.

Input

In the first line, integer n (1 ≤ n ≤ 10^3) - the number of locations discovered.

In next n lines, three integers x, y, c (0 ≤ x, y ≤ 10^4, c ∈ \{0, 1\}) - coordinates of the location and number of civilization it belongs to.

Output

In first line print number of roads that should be built.

In the following lines print all pairs of locations (their 0-based indices) that should be connected with a road.

If it is not possible to build roads such that all restrictions are met, print "Impossible". You should not print the quotation marks.

Example

Input

5
0 0 1
1 0 0
0 1 0
1 1 1
3 2 0


Output

3
1 4
4 2
3 0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using std::make_pair;
using std::pair;
using std::sort;
using std::swap;
const int N = 1e3 + 10;
int n, top, ans[N][2], cnt;
bool incv[N];
struct point {
  int x, y, org;
  bool color;
  inline point(int a = 0, int b = 0) {
    x = a;
    y = b;
  }
  inline const bool operator==(const point &p) const {
    return color == p.color;
  }
  inline const friend double distance(const point &p1, const point &p2) {
    return sqrt(pow(1.0 * (p1.x - p2.x), 2.0) + pow(1.0 * (p1.y - p2.y), 2.0));
  }
} p[N], s[N], ss[N];
struct vector {
  int x, y;
  inline vector(const point &a, const point &b) {
    x = b.x - a.x;
    y = b.y - a.y;
  }
  inline const double length() { return distance(point(0, 0), point(x, y)); }
  inline const friend int cross_product(const vector &v1, const vector &v2) {
    return v1.x * v2.y - v1.y * v2.x;
  }
};
struct triangle {
  point v[3];
  inline triangle(const point &a, const point &b, const point &c) {
    v[0] = a;
    v[1] = b;
    v[2] = c;
  }
  inline const bool difference() {
    int cnt[2];
    cnt[0] = cnt[1] = 0;
    for (int i = 0; i < 3; i++) cnt[v[i].color]++;
    return cnt[0] > cnt[1];
  }
  inline const pair<point, point> same() {
    bool c = difference();
    int s[2], cnt = -1;
    for (int i = 0; i < 3; i++)
      if (v[i].color ^ c) s[++cnt] = i;
    return make_pair(v[s[0]], v[s[1]]);
  }
  inline const bool in(const point &p) {
    int cp1 = cross_product(vector(v[0], v[1]), vector(v[0], p)),
        cp2 = cross_product(vector(v[1], v[2]), vector(v[1], p)),
        cp3 = cross_product(vector(v[2], v[0]), vector(v[2], p));
    return cp1 && cp2 && cp3 && (cp1 > 0) == (cp2 > 0) &&
           (cp1 > 0) == (cp3 > 0);
  }
};
int fa[N];
inline const void init() {
  for (int i = 1; i <= n; i++) fa[i] = i;
}
inline const int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
inline const void Union(int x, int y) { fa[find(x)] = find(y); }
inline const bool operator<(const point &p1, const point &p2) {
  vector v1(p[1], p1), v2(p[1], p2);
  int cp = cross_product(v1, v2);
  if (cp > 0) return true;
  if (!cp && v1.length() < v2.length()) return true;
  return false;
}
inline const void convex_hull() {
  for (int i = 1; i <= n; i++)
    if (p[i].y < p[1].y) swap(p[1], p[i]);
  sort(p + 2, p + n + 1);
  s[++top] = p[1];
  incv[p[1].org] = 1;
  for (int i = 2; i <= n; i++) {
    while (top > 1 &&
           cross_product(vector(s[top - 1], s[top]), vector(s[top], p[i])) <= 0)
      incv[s[top].org] = 0, top--;
    s[++top] = p[i];
    incv[p[i].org] = 1;
  }
}
inline const void divide(triangle t) {
  bool c = t.difference();
  static std::vector<point> same, diff;
  same.clear();
  diff.clear();
  for (int i = 1; i <= n; i++)
    if (t.in(p[i]))
      if (p[i].color ^ c)
        same.push_back(p[i]);
      else
        diff.push_back(p[i]);
  pair<point, point> P = t.same();
  if (diff.empty()) {
    same.push_back(P.first);
    same.push_back(P.second);
    for (int i = 0; i < same.size() - 1; i++)
      for (int u, v, j = i + 1; j < same.size(); j++)
        if (find(u = same[i].org) ^ find(v = same[j].org))
          ans[++cnt][0] = u - 1, ans[cnt][1] = v - 1, Union(u, v);
    return;
  }
  point nxt = diff[0];
  for (int i = 0; i < 2; i++) divide(triangle(t.v[i], t.v[i + 1], nxt));
  divide(triangle(t.v[2], t.v[0], nxt));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].color), p[i].org = i;
  convex_hull();
  init();
  for (int i = 1; i <= top; i++) ss[i] = s[i];
  int tot = std::unique(ss + 1, ss + top + 1) - ss - 1;
  if (tot > 3 || tot == 3 && ss[1].color ^ ss[tot].color)
    return puts("Impossible"), 0;
  int fst = 0, lst = 0;
  bool c = ss[1].color;
  if (tot > 2) {
    for (int i = 1; i <= top; i++) {
      if (s[i].color ^ c && !fst) fst = i;
      if (s[i].color ^ c) lst = i;
    }
    for (int i = fst; i < lst; i++)
      divide(triangle(s[i], s[i + 1], s[lst + 1]));
    for (int i = lst + 1; i < top; i++)
      divide(triangle(s[i], s[i + 1], s[fst]));
    for (int i = 1; i < fst - 1; i++) divide(triangle(s[i], s[i + 1], s[fst]));
    divide(triangle(s[top], s[1], s[fst]));
  }
  if (tot == 2) {
    int pos;
    for (int i = 1; i <= top; i++)
      if (s[i].color ^ c) {
        pos = i;
        break;
      }
    for (int i = pos; i < top; i++)
      divide(triangle(s[pos - 1], s[i], s[i + 1]));
    for (int i = 1; i < pos - 1; i++) divide(triangle(s[i], s[i + 1], s[top]));
  }
  if (tot == 1) {
    point one;
    bool found = 0;
    for (int i = 1; i <= n; i++)
      if (!incv[p[i].org] && p[i].color ^ c) {
        one = p[i];
        found = 1;
        break;
      }
    if (found) {
      for (int i = 1; i < top; i++) divide(triangle(s[i], s[i + 1], one));
      divide(triangle(s[top], s[1], one));
    } else {
      for (int i = 1; i < n; i++)
        for (int j = i + 1; j <= n; j++)
          if (find(i) ^ find(j))
            ans[++cnt][0] = i - 1, ans[cnt][1] = j - 1, Union(i, j);
    }
  }
  printf("%d\n", cnt);
  for (int i = 1; i <= cnt; i++) printf("%d %d\n", ans[i][0], ans[i][1]);
  return 0;
}