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;
const int N = 110000;
const int M = 1100000;
const long long mod = 1e9 + 7;
int n, m;
vector<pair<int, int> > g[N];
int edge[N];
int col[N], dep[N];
int val[N], f[N];
void dfs(int x, int fa) {
  f[x] = fa;
  dep[x] = dep[fa] + 1;
  for (auto p : g[x]) {
    int y = p.first, id = p.second;
    if (y == fa) continue;
    edge[y] = id;
    dfs(y, x);
  }
}
struct query {
  int x, y, w;
  friend bool operator<(query a, query b) { return a.w < b.w; }
} a[N], b[N];
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(make_pair(y, i)), g[y].push_back(make_pair(x, i));
  }
  dfs(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) val[i] = 1e6;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].w);
    b[i] = a[i];
  }
  sort(a + 1, a + 1 + m);
  for (int i = 1; i <= m; i++) {
    int nx = a[i].x, ny = a[i].y;
    while (dep[nx] != dep[ny]) {
      if (dep[nx] > dep[ny])
        val[edge[nx]] = a[i].w, nx = f[nx];
      else
        val[edge[ny]] = a[i].w, ny = f[ny];
    }
    while (nx != ny) {
      val[edge[nx]] = a[i].w, nx = f[nx];
      val[edge[ny]] = a[i].w, ny = f[ny];
    }
  }
  int flag = 0;
  for (int i = 1; i <= m; i++) {
    int nx = b[i].x, ny = b[i].y;
    int mn = 1e9;
    while (dep[nx] != dep[ny]) {
      if (dep[nx] > dep[ny])
        mn = min(mn, val[edge[nx]]), nx = f[nx];
      else
        mn = min(mn, val[edge[ny]]), ny = f[ny];
    }
    while (nx != ny) {
      mn = min(mn, val[edge[nx]]), nx = f[nx];
      mn = min(mn, val[edge[ny]]), ny = f[ny];
    }
    if (mn != b[i].w) flag = 1;
  }
  if (flag)
    printf("-1\n");
  else {
    for (int i = 1; i < n; i++) printf("%d ", val[i]);
  }
  return 0;
}