#include <bits/stdc++.h>
using namespace std;
namespace yspm {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
const long long N = 1e5 + 10;
long long s[N], t[N], minn[N], ans[N];
long long fa[N], dep[N], q, n, cnt, head[N], ed[N];
struct node {
  long long to, next, val;
} e[N << 1];
inline void add(long long u, long long v) {
  e[++cnt].next = head[u];
  e[cnt].to = v;
  return head[u] = cnt, void();
}
inline void dfs(long long x, long long fat) {
  fa[x] = fat;
  dep[x] = dep[fat] + 1;
  for (long long i = head[x]; i; i = e[i].next) {
    long long t = e[i].to;
    if (t == fat) continue;
    ed[t] = (i + 1) >> 1;
    dfs(t, x);
  }
  return;
}
signed main() {
  n = read();
  for (long long i = 1, u, v; i < n; ++i)
    u = read(), v = read(), add(u, v), add(v, u), ans[i] = 1;
  dfs(1, 0);
  q = read();
  for (long long i = 1; i <= q; ++i) {
    s[i] = read();
    t[i] = read();
    minn[i] = read();
    long long x = s[i], y = t[i];
    while (x != y) {
      if (dep[x] < dep[y]) swap(x, y);
      ans[ed[x]] = max(ans[ed[x]], minn[i]);
      x = fa[x];
    }
  }
  for (long long i = 1; i <= q; ++i) {
    long long x = s[i], y = t[i], fl = 0;
    while (x != y) {
      if (dep[x] < dep[y]) swap(x, y);
      if (ans[ed[x]] <= minn[i]) {
        fl = 1;
        break;
      }
      x = fa[x];
    }
    if (!fl) return puts("-1"), 0;
  }
  for (long long i = 1; i < n; ++i) cout << ans[i] << " ";
  puts("");
  return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }

### Prompt

Please create a solution in Cpp 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;
namespace yspm {
inline long long read() {
  long long res = 0, f = 1;
  char k;
  while (!isdigit(k = getchar()))
    if (k == '-') f = -1;
  while (isdigit(k)) res = res * 10 + k - '0', k = getchar();
  return res * f;
}
const long long N = 1e5 + 10;
long long s[N], t[N], minn[N], ans[N];
long long fa[N], dep[N], q, n, cnt, head[N], ed[N];
struct node {
  long long to, next, val;
} e[N << 1];
inline void add(long long u, long long v) {
  e[++cnt].next = head[u];
  e[cnt].to = v;
  return head[u] = cnt, void();
}
inline void dfs(long long x, long long fat) {
  fa[x] = fat;
  dep[x] = dep[fat] + 1;
  for (long long i = head[x]; i; i = e[i].next) {
    long long t = e[i].to;
    if (t == fat) continue;
    ed[t] = (i + 1) >> 1;
    dfs(t, x);
  }
  return;
}
signed main() {
  n = read();
  for (long long i = 1, u, v; i < n; ++i)
    u = read(), v = read(), add(u, v), add(v, u), ans[i] = 1;
  dfs(1, 0);
  q = read();
  for (long long i = 1; i <= q; ++i) {
    s[i] = read();
    t[i] = read();
    minn[i] = read();
    long long x = s[i], y = t[i];
    while (x != y) {
      if (dep[x] < dep[y]) swap(x, y);
      ans[ed[x]] = max(ans[ed[x]], minn[i]);
      x = fa[x];
    }
  }
  for (long long i = 1; i <= q; ++i) {
    long long x = s[i], y = t[i], fl = 0;
    while (x != y) {
      if (dep[x] < dep[y]) swap(x, y);
      if (ans[ed[x]] <= minn[i]) {
        fl = 1;
        break;
      }
      x = fa[x];
    }
    if (!fl) return puts("-1"), 0;
  }
  for (long long i = 1; i < n; ++i) cout << ans[i] << " ";
  puts("");
  return 0;
}
}  // namespace yspm
signed main() { return yspm::main(); }

```