Problem: You are given a tree with n nodes. You have to write non-negative integers on its edges so that the following condition would be satisfied:

For every two nodes i, j, look at the path between them and count the sum of numbers on the edges of this path. Write all obtained sums on the blackboard. Then every integer from 1 to ⌊ (2n^2)/(9) ⌋ has to be written on the blackboard at least once. 

It is guaranteed that such an arrangement exists.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of nodes.

Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), meaning that there is an edge between nodes u and v. It is guaranteed that these edges form a tree.

Output

Output n-1 lines, each of form u v x (0 ≤ x ≤ 10^6), which will mean that you wrote number x on the edge between u, v. 

Set of edges (u, v) has to coincide with the set of edges of the input graph, but you can output edges in any order. You can also output ends of edges in an order different from the order in input.

Examples

Input


3
2 3
2 1


Output


3 2 1
1 2 2


Input


4
2 4
2 3
2 1


Output


4 2 1
3 2 2
1 2 3


Input


5
1 2
1 3
1 4
2 5


Output


2 1 1
5 2 1
3 1 3
4 1 6

Note

In the first example, distance between nodes 1 and 2 is equal to 2, between nodes 2 and 3 to 1, between 1 and 3 to 3.

In the third example, numbers from 1 to 9 (inclusive) will be written on the blackboard, while we need just from 1 to 5 to pass the test.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[1000010];
long long inv[1000010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
long long sita[1010];
long long oya[1010];
vector<long long> G[1010], H1[1010], H2[1010];
long long dfs1(long long r, long long p) {
  oya[r] = p;
  sita[r] = 1;
  for (auto v : G[r])
    if (v != p) {
      sita[r] += dfs1(v, r);
    }
  return sita[r];
}
long long dep[1011];
void dfs2(long long r, long long p, long long d = 1) {
  oya[r] = p;
  dep[r] = d;
  for (auto v : G[r])
    if (v != p) {
      dfs2(v, r, d + 1);
      H1[r].push_back(v);
    }
}
void dfs3(long long r, long long p, long long d = 1) {
  oya[r] = p;
  dep[r] = d;
  for (auto v : G[r])
    if (v != p) {
      dfs3(v, r, d + 1);
      H2[r].push_back(v);
    }
}
long long dp[1010][1010];
long long mae[1010][1011];
long long wei[1010];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long y, yy;
    cin >> y >> yy;
    G[y].push_back(yy);
    G[yy].push_back(y);
  }
  if (n == 1) {
    return 0;
  }
  if (n == 2) {
    cout << "1 2 1" << endl;
    return 0;
  }
  dfs1(1, 1);
  long long root = -1;
  for (long long i = 2; i <= n; i++) {
    long long wa = n - 1;
    bool ok = true;
    for (auto v : G[i])
      if (v != oya[i]) {
        if (!(sita[v] <= n / 2)) ok = 0;
        wa -= sita[v];
      }
    if (!ok) continue;
    if (wa <= n / 2) root = i;
  }
  if (root == -1) root = 1;
  dfs1(root, root);
  vector<pair<long long, long long> > ve;
  for (auto v : G[root]) ve.push_back({v, sita[v]});
  long long m = ve.size();
  dp[0][0] = 1;
  mae[0][0] = -1;
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j <= n; j++)
      if (dp[i][j]) {
        dp[i + 1][j] = 1;
        mae[i + 1][j] = mae[i][j];
        dp[i + 1][j + ve[i].second] = 1;
        mae[i + 1][j + ve[i].second] = i;
      }
  }
  vector<long long> V1, V2;
  for (long long j = n / 2; j >= 0; j--)
    if (dp[m][j]) {
      long long imaval = j;
      long long imaind = m;
      while (1) {
        V1.push_back(ve[mae[imaind][imaval]].first);
        long long ee = imaval;
        imaval -= ve[mae[imaind][imaval]].second;
        imaind = mae[imaind][ee];
        if (mae[imaind][imaval] == -1) break;
      }
      break;
    }
  for (auto v : ve) {
    bool bo = 0;
    for (auto w : V1)
      if (v.first == w) {
        bo = 1;
        break;
      }
    if (!bo) V2.push_back(v.first);
  }
  for (auto v : V1) dfs2(v, root), H1[root].push_back(v);
  for (auto v : V2) dfs3(v, root), H2[root].push_back(v);
  long long h1 = 0, h2 = 0;
  for (long long i = 1; i <= n; i++) h1 += H1[i].size();
  for (long long i = 1; i <= n; i++) h2 += H2[i].size();
  assert(h1 + h2 == n - 1);
  ve.clear();
  for (long long i = 1; i <= n; i++)
    for (auto v : H1[i]) ve.push_back({dep[v], v});
  sort(ve.begin(), ve.end());
  for (long long i = 0; i < ve.size(); i++) {
    wei[ve[i].second] = (i + 1) * (h2 + 1);
  }
  ve.clear();
  for (long long i = 1; i <= n; i++)
    for (auto v : H2[i]) ve.push_back({dep[v], v});
  sort(ve.begin(), ve.end());
  for (long long i = 0; i < ve.size(); i++) {
    wei[ve[i].second] = i + 1;
  }
  for (long long i = 1; i <= n; i++)
    if (i != root) {
      cout << i << " " << oya[i] << " " << wei[i] - wei[oya[i]] << endl;
    }
  for (long long i = 1; i <= n; i++)
    if (i != root) {
    }
  return 0;
}