#include <bits/stdc++.h>
using namespace std;
int n, m;
int U[5005], V[5005];
struct emm {
  int a, b, x;
} a[5005];
bool cmp(emm x, emm y) { return x.x < y.x; }
struct Edge {
  int v;
  Edge *next;
} pool[5005 << 1], *h[5005];
int fa[5005], dep[5005], ans[5005];
int tot;
void addEdge(int u, int v) {
  Edge *p = &pool[tot++];
  p->v = v;
  p->next = h[u];
  h[u] = p;
}
void dfs(int u) {
  for (Edge *p = h[u]; p; p = p->next) {
    int v = p->v;
    if (v == fa[u]) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    dfs(v);
  }
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    u = fa[u];
  }
  if (u == v) return u;
  while (fa[u] != fa[v]) {
    u = fa[u];
    v = fa[v];
  }
  return fa[u];
}
void update(int u, int v, int x) {
  if (u == v) return;
  while (u != v) {
    ans[u] = x;
    u = fa[u];
  }
}
int check(int u, int v) {
  int res = 0x3f3f3f3f;
  if (u == v) return res;
  while (u != v) {
    res = min(res, ans[u]);
    u = fa[u];
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &U[i], &V[i]);
    addEdge(U[i], V[i]);
    addEdge(V[i], U[i]);
  }
  dep[1] = 1;
  dfs(1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &a[i].a, &a[i].b, &a[i].x);
  sort(a + 1, a + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    int t = LCA(a[i].a, a[i].b);
    update(a[i].a, t, a[i].x);
    update(a[i].b, t, a[i].x);
  }
  for (int i = 1; i <= m; i++) {
    int t = LCA(a[i].a, a[i].b);
    if (min(check(a[i].a, t), check(a[i].b, t)) != a[i].x) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (dep[U[i]] < dep[V[i]]) swap(U[i], V[i]);
    printf("%d ", max(ans[U[i]], 1));
  }
  printf("\n");
  return 0;
}

### Prompt

Generate 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;
int n, m;
int U[5005], V[5005];
struct emm {
  int a, b, x;
} a[5005];
bool cmp(emm x, emm y) { return x.x < y.x; }
struct Edge {
  int v;
  Edge *next;
} pool[5005 << 1], *h[5005];
int fa[5005], dep[5005], ans[5005];
int tot;
void addEdge(int u, int v) {
  Edge *p = &pool[tot++];
  p->v = v;
  p->next = h[u];
  h[u] = p;
}
void dfs(int u) {
  for (Edge *p = h[u]; p; p = p->next) {
    int v = p->v;
    if (v == fa[u]) continue;
    dep[v] = dep[u] + 1;
    fa[v] = u;
    dfs(v);
  }
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  while (dep[u] > dep[v]) {
    u = fa[u];
  }
  if (u == v) return u;
  while (fa[u] != fa[v]) {
    u = fa[u];
    v = fa[v];
  }
  return fa[u];
}
void update(int u, int v, int x) {
  if (u == v) return;
  while (u != v) {
    ans[u] = x;
    u = fa[u];
  }
}
int check(int u, int v) {
  int res = 0x3f3f3f3f;
  if (u == v) return res;
  while (u != v) {
    res = min(res, ans[u]);
    u = fa[u];
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &U[i], &V[i]);
    addEdge(U[i], V[i]);
    addEdge(V[i], U[i]);
  }
  dep[1] = 1;
  dfs(1);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &a[i].a, &a[i].b, &a[i].x);
  sort(a + 1, a + m + 1, cmp);
  for (int i = 1; i <= m; i++) {
    int t = LCA(a[i].a, a[i].b);
    update(a[i].a, t, a[i].x);
    update(a[i].b, t, a[i].x);
  }
  for (int i = 1; i <= m; i++) {
    int t = LCA(a[i].a, a[i].b);
    if (min(check(a[i].a, t), check(a[i].b, t)) != a[i].x) {
      printf("-1\n");
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (dep[U[i]] < dep[V[i]]) swap(U[i], V[i]);
    printf("%d ", max(ans[U[i]], 1));
  }
  printf("\n");
  return 0;
}

```