#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x *= f, void();
}
const int N = 5000 + 5, INF = 1e9;
int n, m, U[N], V[N], W[N], fa[N], pre[N], dep[N], qu[N], qv[N], qw[N];
vector<int> g[N];
void dfs(int u) {
  for (int e : g[u]) {
    int v = (u == U[e] ? V[e] : U[e]);
    if (v == fa[u]) continue;
    fa[v] = u, dep[v] = dep[u] + 1, pre[v] = e, dfs(v);
  }
}
inline void ckmax(int u, int v, int w) {
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    W[pre[u]] = max(W[pre[u]], w);
    u = fa[u];
  }
  while (u != v) {
    W[pre[u]] = max(W[pre[u]], w);
    u = fa[u];
    W[pre[v]] = max(W[pre[v]], w);
    v = fa[v];
  }
}
inline int Min(int u, int v) {
  int ans = INF;
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    ans = min(ans, W[pre[u]]);
    u = fa[u];
  }
  while (u != v) {
    ans = min(ans, W[pre[u]]);
    u = fa[u];
    ans = min(ans, W[pre[v]]);
    v = fa[v];
  }
  return ans;
}
int main() {
  read(n);
  for (int i = 1; i < n; i++) {
    read(U[i]), read(V[i]);
    g[U[i]].push_back(i);
    g[V[i]].push_back(i);
    W[i] = 1;
  }
  dfs(1);
  read(m);
  for (int i = 1; i <= m; i++) {
    read(qu[i]), read(qv[i]), read(qw[i]);
    ckmax(qu[i], qv[i], qw[i]);
  }
  for (int i = 1; i <= m; i++) {
    if (Min(qu[i], qv[i]) != qw[i]) {
      puts("-1");
      exit(0);
    }
  }
  for (int i = 1; i < n; i++) printf("%d\n", W[i]);
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to 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;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x *= f, void();
}
const int N = 5000 + 5, INF = 1e9;
int n, m, U[N], V[N], W[N], fa[N], pre[N], dep[N], qu[N], qv[N], qw[N];
vector<int> g[N];
void dfs(int u) {
  for (int e : g[u]) {
    int v = (u == U[e] ? V[e] : U[e]);
    if (v == fa[u]) continue;
    fa[v] = u, dep[v] = dep[u] + 1, pre[v] = e, dfs(v);
  }
}
inline void ckmax(int u, int v, int w) {
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    W[pre[u]] = max(W[pre[u]], w);
    u = fa[u];
  }
  while (u != v) {
    W[pre[u]] = max(W[pre[u]], w);
    u = fa[u];
    W[pre[v]] = max(W[pre[v]], w);
    v = fa[v];
  }
}
inline int Min(int u, int v) {
  int ans = INF;
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    ans = min(ans, W[pre[u]]);
    u = fa[u];
  }
  while (u != v) {
    ans = min(ans, W[pre[u]]);
    u = fa[u];
    ans = min(ans, W[pre[v]]);
    v = fa[v];
  }
  return ans;
}
int main() {
  read(n);
  for (int i = 1; i < n; i++) {
    read(U[i]), read(V[i]);
    g[U[i]].push_back(i);
    g[V[i]].push_back(i);
    W[i] = 1;
  }
  dfs(1);
  read(m);
  for (int i = 1; i <= m; i++) {
    read(qu[i]), read(qv[i]), read(qw[i]);
    ckmax(qu[i], qv[i], qw[i]);
  }
  for (int i = 1; i <= m; i++) {
    if (Min(qu[i], qv[i]) != qw[i]) {
      puts("-1");
      exit(0);
    }
  }
  for (int i = 1; i < n; i++) printf("%d\n", W[i]);
  return 0;
}

```