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 long long mod = 1000000007;
const long long INF = (1e18);
struct edge {
  int v, nxt, val;
} e[((5000) + 5) * 2];
int head[((5000) + 5)], ef[((5000) + 5)], en = 0, n;
int dis[((5000) + 5)], fa[((5000) + 5)][20];
int flag = 1;
void adde(int u, int v) {
  e[en].v = v;
  e[en].nxt = head[u];
  e[en].val = 0;
  head[u] = en;
  en++;
}
void dfs0(int x, int pre) {
  for (int i = head[x]; i != -1; i = e[i].nxt) {
    int v = e[i].v;
    if (v == pre) continue;
    ef[v] = (i ^ 1);
    fa[v][0] = x;
    dis[v] = dis[x] + 1;
    dfs0(v, x);
  }
}
void init() {
  dfs0(1, -1);
  for (int i = 1; (1 << i) < n; i++) {
    for (int j = 1; j <= n; j++) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
    }
  }
}
bool col(int x, int rt, int w) {
  bool flag1 = 0;
  while (x != rt) {
    int ei = ef[x];
    if (!e[ei].val) {
      e[ei].val = w;
      e[ei ^ 1].val = w;
      flag1 = 1;
    } else {
      if (e[ei].val == w) flag1 = 1;
    }
    x = e[ei].v;
  }
  return flag1;
}
int lca(int a, int b) {
  if (dis[a] > dis[b]) swap(a, b);
  int d = dis[b] - dis[a];
  for (int i = 0; (1 << i) <= d; i++) {
    if (d & (1 << i)) {
      b = fa[b][i];
    }
  }
  if (a != b) {
    for (int i = log2(n); i >= 0; i--) {
      if (fa[a][i] != fa[b][i]) {
        a = fa[a][i];
        b = fa[b][i];
      }
    }
    a = fa[a][0];
  }
  return a;
}
struct qre {
  int ai, bi, gi;
} q[((5000) + 5)];
bool cmp(qre a, qre b) { return a.gi > b.gi; }
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; i++) {
    int v, u;
    scanf("%d%d", &u, &v);
    adde(u, v);
    adde(v, u);
  }
  int m;
  scanf("%d", &m);
  init();
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &q[i].ai, &q[i].bi, &q[i].gi);
  sort(q + 1, q + 1 + m, cmp);
  for (int i = 1; i <= m; i++) {
    int rt = lca(q[i].ai, q[i].bi);
    bool fg1 = col(q[i].ai, rt, q[i].gi);
    bool fg2 = col(q[i].bi, rt, q[i].gi);
    if (!fg1 && !fg2) flag = 0;
  }
  if (flag) {
    for (int i = 1; i < en; i += 2) {
      printf("%d%c", (!e[i].val) ? 1 : e[i].val, i == en - 1 ? '\n' : ' ');
    }
  } else {
    printf("-1\n");
  }
  return 0;
}