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>
#pragma GCC optimize( \
    "Ofast,no-stack-protector,unroll-loops,no-stack-protector,fast-math")
using namespace std;
const int N = 5e3 + 5, M = 2 * N + 5;
int head[N], nxt[M], to[M], mx[M], ne, n, m;
void init() {
  ne = 0;
  memset(head, -1, n * sizeof head[0]);
}
void addEdge(int f, int t) {
  to[ne] = t;
  nxt[ne] = head[f];
  head[f] = ne++;
}
void addBiEdge(int a, int b) {
  addEdge(a, b);
  addEdge(b, a);
}
int par[N], parEdge[N], dep[N];
void dfs(int u = 0, int p = -1, int d = 0) {
  dep[u] = d, par[u] = p;
  for (int e = head[u]; ~e; e = nxt[e]) {
    int v = to[e];
    if (v == p) continue;
    dfs(v, u, d + 1);
    parEdge[v] = e;
  }
}
void upd(int &u, int &v, int &e) {
  if (dep[u] > dep[v]) {
    e = parEdge[u];
    u = par[u];
  } else {
    e = parEdge[v];
    v = par[v];
  }
}
void setVal(int u, int v, int val) {
  int e;
  while (u != v) {
    upd(u, v, e);
    mx[e] = mx[e ^ 1] = max({mx[e], mx[e ^ 1], val});
  }
}
int getMin(int u, int v) {
  int ret = 1e6, e;
  while (u != v) {
    upd(u, v, e);
    ret = min({ret, mx[e], mx[e ^ 1]});
  }
  return ret;
}
int u[N], v[N], val[N];
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    addBiEdge(--u, --v);
  }
  dfs();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d", u + i, v + i, val + i);
    setVal(--u[i], --v[i], val[i]);
  }
  for (int i = 0; i < 2 * n; ++i)
    if (!mx[i]) mx[i] = 1e6;
  for (int i = 0; i < q; ++i)
    if (getMin(u[i], v[i]) != val[i]) exit(puts("-1"));
  for (int i = 0; i < 2 * n - 2; i += 2)
    printf("%d%c", mx[i], " \n"[i + 1 == 2 * n - 2]);
}