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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  if (a == 0ll) {
    return 0ll;
  }
  a %= MOD;
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return ans;
}
void Pv(const vector<int> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%d", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
void Pvl(const vector<long long> &V) {
  int Len = int(V.size());
  for (int i = 0; i < Len; ++i) {
    printf("%lld", V[i]);
    if (i != Len - 1) {
      printf(" ");
    } else {
      printf("\n");
    }
  }
}
inline long long readll() {
  long long tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
inline int readint() {
  int tmp = 0, fh = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar();
  return tmp * fh;
}
const int maxn = 5010;
const int inf = 0x3f3f3f3f;
int n;
int id[maxn][maxn];
int fa[maxn];
struct node {
  int f, t, w;
  bool operator<(const node &b) const { return w > b.w; }
} e[maxn];
int ans[maxn];
int m;
std::vector<int> son[maxn];
void dfs(int x, int pre) {
  for (auto y : son[x]) {
    if (y != pre) {
      fa[y] = x;
      dfs(y, x);
    }
  }
}
int main() {
  n = readint();
  for (int i = 2; i <= n; i++) {
    int x, y;
    x = readint();
    y = readint();
    son[x].push_back(y);
    son[y].push_back(x);
    id[x][y] = id[y][x] = i;
  }
  m = readint();
  for (int i = 1; i <= m; i++) {
    e[i].f = readint();
    e[i].t = readint();
    e[i].w = readint();
  }
  sort(e + 1, e + 1 + m);
  int flag = 1;
  for (int i = 1; i <= m; i++) {
    dfs(e[i].f, 0);
    int now = e[i].t;
    int bj = 0;
    do {
      int j = id[now][fa[now]];
      if (ans[j] <= e[i].w) {
        bj = 1;
        ans[j] = e[i].w;
      }
      now = fa[now];
    } while (now != e[i].f);
    if (!bj) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (int i = 2; i <= n; i++) {
      printf("%d%c", max(ans[i], 1), i == n ? '\n' : ' ');
    }
  } else {
    printf("-1\n");
  }
  return 0;
}