#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e6 + 3;
void rd(int &x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
void pr(int x) { printf("%d\n", x); }
pair<pair<int, int>, int> pa[5005];
vector<pair<int, int> > g[5005];
int eg[5005];
bool dfs(int fa, int s, int t, int w) {
  bool bj = 0;
  if (s == t) return 1;
  for (auto u : g[s]) {
    int x = u.first;
    if (x != fa)
      if (dfs(s, x, t, w)) eg[u.second] = max(eg[u.second], w), bj = 1;
  }
  return bj;
}
bool check(int fa, int s, int t, int &ans) {
  bool bj = 0;
  if (s == t) return 1;
  for (auto u : g[s]) {
    int x = u.first;
    if (x != fa)
      if (check(s, x, t, ans)) ans = min(eg[u.second], ans), bj = 1;
  }
  return bj;
}
int main() {
  int n, m;
  rd(n);
  for (int i = (0); i < n - 1; i++) {
    int a, b;
    rd(a), rd(b);
    g[a].push_back({b, i});
    g[b].push_back({a, i});
  }
  rd(m);
  for (int i = (0); i < m; i++) {
    int a, b, c;
    rd(a), rd(b), rd(c);
    pa[i] = {{a, b}, c};
    dfs(0, a, b, c);
  }
  for (int i = (0); i < m; i++) {
    int ans = 0x3f3f3f3f;
    int s = pa[i].first.first, t = pa[i].first.second, w = pa[i].second;
    check(0, s, t, ans);
    if (ans != w) return 0 * puts("-1");
  }
  for (int i = (0); i < n - 1; i++) {
    if (eg[i])
      printf("%d ", eg[i]);
    else
      printf("1 ");
  }
}

### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e6 + 3;
void rd(int &x) {
  int f = 1;
  x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
void pr(int x) { printf("%d\n", x); }
pair<pair<int, int>, int> pa[5005];
vector<pair<int, int> > g[5005];
int eg[5005];
bool dfs(int fa, int s, int t, int w) {
  bool bj = 0;
  if (s == t) return 1;
  for (auto u : g[s]) {
    int x = u.first;
    if (x != fa)
      if (dfs(s, x, t, w)) eg[u.second] = max(eg[u.second], w), bj = 1;
  }
  return bj;
}
bool check(int fa, int s, int t, int &ans) {
  bool bj = 0;
  if (s == t) return 1;
  for (auto u : g[s]) {
    int x = u.first;
    if (x != fa)
      if (check(s, x, t, ans)) ans = min(eg[u.second], ans), bj = 1;
  }
  return bj;
}
int main() {
  int n, m;
  rd(n);
  for (int i = (0); i < n - 1; i++) {
    int a, b;
    rd(a), rd(b);
    g[a].push_back({b, i});
    g[b].push_back({a, i});
  }
  rd(m);
  for (int i = (0); i < m; i++) {
    int a, b, c;
    rd(a), rd(b), rd(c);
    pa[i] = {{a, b}, c};
    dfs(0, a, b, c);
  }
  for (int i = (0); i < m; i++) {
    int ans = 0x3f3f3f3f;
    int s = pa[i].first.first, t = pa[i].first.second, w = pa[i].second;
    check(0, s, t, ans);
    if (ans != w) return 0 * puts("-1");
  }
  for (int i = (0); i < n - 1; i++) {
    if (eg[i])
      printf("%d ", eg[i]);
    else
      printf("1 ");
  }
}

```