Problem: We had a really tough time generating tests for problem D. In order to prepare strong tests, we had to solve the following problem.

Given an undirected labeled tree consisting of n vertices, find a set of segments such that:

  1. both endpoints of each segment are integers from 1 to 2n, and each integer from 1 to 2n should appear as an endpoint of exactly one segment; 
  2. all segments are non-degenerate; 
  3. for each pair (i, j) such that i ≠ j, i ∈ [1, n] and j ∈ [1, n], the vertices i and j are connected with an edge if and only if the segments i and j intersect, but neither segment i is fully contained in segment j, nor segment j is fully contained in segment i. 



Can you solve this problem too?

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of vertices in the tree.

Then n - 1 lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) denoting the endpoints of the i-th edge.

It is guaranteed that the given graph is a tree.

Output

Print n pairs of integers, the i-th pair should contain two integers l_i and r_i (1 ≤ l_i < r_i ≤ 2n) — the endpoints of the i-th segment. All 2n integers you print should be unique.

It is guaranteed that the answer always exists.

Examples

Input


6
1 2
1 3
3 4
3 5
2 6


Output


9 12
7 10
3 11
1 5
2 4
6 8


Input


1


Output


1 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  int x;
  int y, z;
  pa3(int x = 0, int y = 0, int 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:
  int x;
  int y, z, w;
  pa4(int x = 0, int y = 0, int z = 0, int 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:
  int x, y;
  pa2(int x = 0, int 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(int i) {
  ostringstream s;
  s << i;
  return s.str();
}
int gcd(int v, int b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
int mod;
int extgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<int, int> operator+(const pair<int, int> &l, const pair<int, int> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<int, int> operator-(const pair<int, int> &l, const pair<int, int> &r) {
  return {l.first - r.first, l.second - r.second};
}
int pr[10000100];
int inv[10000010];
int beki(int wa, int rr, int 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;
}
int comb(int nn, int rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  int r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(int ert) {
  pr[0] = 1;
  for (int i = 1; i <= ert; i++) {
    pr[i] = ((long long)pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (int i = ert - 1; i >= 0; i--) {
    inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
  }
}
vector<int> G[500020];
vector<pa3> ve[500020];
int sita[500020], ch[500020];
void yaru(vector<pa3> &ve, int &mi, int &ko, vector<pa3> &ve2, int mi2,
          int ko2) {
  if (ve.size() < ve2.size()) {
    swap(ve, ve2), swap(mi, mi2), swap(ko, ko2);
  }
  int n = ve.size();
  int m = ve2.size();
  int cnt = 0;
  for (auto &v : ve2) {
    bool bo = 0;
    if (v.x < mi2 + ko2) {
      v.x = mi - ko2 + v.x - mi2;
      bo = 1;
      cnt++;
    } else
      v.x = mi + 2 * n + v.x - mi2 - ko2;
    v.y = mi + 2 * n + v.y - mi2 - ko2;
    ve.push_back(v);
    if (bo) swap(ve[ko + cnt - 1], ve.back());
  }
  mi -= ko2;
  ko += ko2;
}
void dfs(int r, int p) {
  int cnt = 0;
  for (auto v : G[r])
    if (v != p) {
      dfs(v, r);
      cnt++;
    }
  if (cnt == 0) {
    ve[r].push_back((pa3){0, 1, r});
    sita[r] = 0, ch[r] = 1;
    return;
  }
  int cnt2 = 0;
  vector<pa3> ima;
  int sai, ko;
  for (auto v : G[r])
    if (v != p) {
      cnt2++;
      if (cnt2 == 1) {
        swap(ima, ve[v]);
        sai = sita[v];
        ko = 1;
      } else {
        yaru(ima, sai, ko, ve[v], sita[v], 1);
      }
    }
  for (int i = 0; i < ko; i++)
    if (ima[i].x < sai + ko) ima[i].x--;
  ima.push_back((pa3){sai - 2, sai + ko - 1, r});
  swap(ima[0], ima.back());
  sita[r] = sai - 2, ch[r] = ko + 1;
  swap(ve[r], ima);
}
int ans[500020][2];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int y, yy;
    scanf("%d %d", &y, &yy);
    G[y].push_back(yy);
    G[yy].push_back(y);
  }
  dfs(n, n);
  for (auto v : ve[n]) {
    ans[v.z][0] = v.x - sita[n] + 1;
    ans[v.z][1] = v.y - sita[n] + 1;
  }
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", ans[i][0], ans[i][1]);
  }
  return 0;
}