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 maxn = 5000 + 10;
int n;
int m;
int lg[maxn];
int depth[maxn];
int fa[maxn][22];
int edgenum[maxn];
int w[maxn];
struct Edge {
  int to, next, idx;
} e[maxn << 1];
int head[maxn];
int en;
void add(int from, int to, int idx) {
  e[en].next = head[from];
  e[en].to = to;
  e[en].idx = idx;
  head[from] = en;
  ++en;
}
void dfs(int u, int pre) {
  depth[u] = depth[pre] + 1;
  fa[u][0] = pre;
  for (int i = 1; i <= lg[depth[u] - 1]; ++i) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    int id = e[i].idx;
    if (v == pre) continue;
    edgenum[v] = id;
    dfs(v, u);
  }
}
int LCA(int p, int q) {
  if (depth[p] < depth[q]) swap(p, q);
  while (depth[p] > depth[q]) {
    p = fa[p][lg[depth[p] - depth[q]]];
  }
  if (p == q) return p;
  for (int i = lg[depth[p] - 1]; i >= 0; --i) {
    if (fa[p][i] != fa[q][i]) {
      p = fa[p][i];
      q = fa[q][i];
    }
  }
  return fa[p][0];
}
void modify(int u, int v, int val) {
  int lca = LCA(u, v);
  while (u != lca) {
    w[edgenum[u]] = max(w[edgenum[u]], val);
    u = fa[u][0];
  }
  while (v != lca) {
    w[edgenum[v]] = max(w[edgenum[v]], val);
    v = fa[v][0];
  }
}
int que(int u, int v) {
  int mi = 1000001;
  int lca = LCA(u, v);
  while (u != lca) {
    mi = min(mi, w[edgenum[u]]);
    u = fa[u][0];
  }
  while (v != lca) {
    mi = min(mi, w[edgenum[v]]);
    v = fa[v][0];
  }
  return mi;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    add(u, v, i);
    add(v, u, i);
  }
  scanf("%d", &m);
  vector<vector<int>> query(m + 1, vector<int>(3, 0));
  for (int i = 1; i <= m; ++i) {
    scanf("%d %d %d", &query[i][0], &query[i][1], &query[i][2]);
  }
  lg[1] = 0;
  lg[2] = 1;
  for (int i = 3; i <= n; ++i) {
    if (i == (1 << (lg[i - 1] + 1))) {
      lg[i] = lg[i - 1] + 1;
    } else {
      lg[i] = lg[i - 1];
    }
  }
  dfs(1, 0);
  for (int i = 1; i <= m; ++i) {
    int u = query[i][0];
    int v = query[i][1];
    int val = query[i][2];
    modify(u, v, val);
  }
  for (int i = 1; i <= m; ++i) {
    int u = query[i][0];
    int v = query[i][1];
    int val = query[i][2];
    int q = que(u, v);
    if (q != val) {
      printf("-1");
      return 0;
    }
  }
  for (int i = 1; i <= n - 1; ++i) {
    if (w[i] == 0) w[i] = 1;
    printf("%d ", w[i]);
  }
  return 0;
}