Problem: There are n railway stations in Berland. They are connected to each other by n-1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n-1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 1 to 10^6 inclusive.

You asked m passengers some questions: the j-th one told you three values:

  * his departure station a_j; 
  * his arrival station b_j; 
  * minimum scenery beauty along the path from a_j to b_j (the train is moving along the shortest path from a_j to b_j). 



You are planning to update the map and set some value f_i on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.

Print any valid set of values f_1, f_2, ..., f_{n-1}, which the passengers' answer is consistent with or report that it doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of railway stations in Berland.

The next n-1 lines contain descriptions of the railway sections: the i-th section description is two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), where x_i and y_i are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer m (1 ≤ m ≤ 5000) — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers a_j, b_j and g_j (1 ≤ a_j, b_j ≤ n; a_j ≠ b_j; 1 ≤ g_j ≤ 10^6) — the departure station, the arrival station and the minimum scenery beauty along his path.

Output

If there is no answer then print a single integer -1.

Otherwise, print n-1 integers f_1, f_2, ..., f_{n-1} (1 ≤ f_i ≤ 10^6), where f_i is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

Input


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


Output


5 3 5


Input


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


Output


5 3 1 2 1 


Input


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


Output


-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct zh {
  int u, v;
  int w;
} a[5005];
zh b[5005];
vector<int> mmap[5005];
int fa[5005];
int dep[5005];
int qz[5005][5005];
void dfs(int nnow, int ffa) {
  int len = mmap[nnow].size();
  int i = 0;
  while (i < len) {
    if (mmap[nnow][i] != ffa) {
      int tt = mmap[nnow][i];
      fa[tt] = nnow;
      dep[tt] = dep[nnow] + 1;
      dfs(tt, nnow);
    }
    i++;
  }
}
void gs(int u, int v, int w) {
  while (dep[u] > dep[v]) {
    qz[fa[u]][u] = max(qz[fa[u]][u], w);
    u = fa[u];
  }
  while (dep[u] < dep[v]) {
    qz[fa[v]][v] = max(qz[fa[v]][v], w);
    v = fa[v];
  }
  while (v != u) {
    qz[fa[u]][u] = max(qz[fa[u]][u], w);
    u = fa[u];
    qz[fa[v]][v] = max(qz[fa[v]][v], w);
    v = fa[v];
  }
}
int check(int u, int v, int w) {
  int mmin = 99999999;
  while (dep[u] > dep[v]) {
    mmin = min(qz[fa[u]][u], mmin);
    u = fa[u];
  }
  while (dep[u] < dep[v]) {
    mmin = min(qz[fa[v]][v], mmin);
    v = fa[v];
  }
  while (v != u) {
    mmin = min(qz[fa[u]][u], mmin);
    u = fa[u];
    mmin = min(qz[fa[v]][v], mmin);
    v = fa[v];
  }
  if (mmin == w) return 1;
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  int i = 1;
  while (i < n) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[i].u = x;
    a[i].v = y;
    mmap[x].push_back(y);
    mmap[y].push_back(x);
    i++;
  }
  dep[1] = 1;
  dfs(1, 1);
  int m;
  scanf("%d", &m);
  i = 1;
  while (i <= m) {
    scanf("%d%d%d", &b[i].u, &b[i].v, &b[i].w);
    gs(b[i].u, b[i].v, b[i].w);
    i++;
  }
  int f1 = 1;
  i = 1;
  while (i <= m) {
    if (check(b[i].u, b[i].v, b[i].w) == 0) {
      f1 = 0;
      break;
    }
    i++;
  }
  if (f1 == 0)
    cout << -1 << endl;
  else {
    i = 1;
    while (i < n) {
      int x = a[i].u;
      int y = a[i].v;
      int sum = qz[x][y] + qz[y][x];
      if (sum == 0) sum++;
      cout << sum << " ";
      i++;
    }
  }
  return 0;
}