Ali is Hamed's little brother and tomorrow is his birthday. Hamed wants his brother to earn his gift so he gave him a hard programming problem and told him if he can successfully solve it, he'll get him a brand new laptop. Ali is not yet a very talented programmer like Hamed and although he usually doesn't cheat but this time is an exception. It's about a brand new laptop. So he decided to secretly seek help from you. Please solve this problem for Ali. 

An n-vertex weighted rooted tree is given. Vertex number 1 is a root of the tree. We define d(u, v) as the sum of edges weights on the shortest path between vertices u and v. Specifically we define d(u, u) = 0. Also let's define S(v) for each vertex v as a set containing all vertices u such that d(1, u) = d(1, v) + d(v, u). Function f(u, v) is then defined using the following formula:

<image>

The goal is to calculate f(u, v) for each of the q given pair of vertices. As the answer can be rather large it's enough to print it modulo 109 + 7.

Input

In the first line of input an integer n (1 ≤ n ≤ 105), number of vertices of the tree is given.

In each of the next n - 1 lines three space-separated integers ai, bi, ci (1 ≤ ai, bi ≤ n, 1 ≤ ci ≤ 109) are given indicating an edge between ai and bi with weight equal to ci.

In the next line an integer q (1 ≤ q ≤ 105), number of vertex pairs, is given.

In each of the next q lines two space-separated integers ui, vi (1 ≤ ui, vi ≤ n) are given meaning that you must calculate f(ui, vi).

It is guaranteed that the given edges form a tree.

Output

Output q lines. In the i-th line print the value of f(ui, vi) modulo 109 + 7.

Examples

Input

5
1 2 1
4 3 1
3 5 1
1 3 1
5
1 1
1 5
2 4
2 1
3 5


Output

10
1000000005
1000000002
23
1000000002


Input

8
1 2 100
1 3 20
2 4 2
2 5 1
3 6 1
3 7 2
6 8 5
6
1 8
2 3
5 8
2 6
4 7
6 1


Output

999968753
49796
999961271
999991235
999958569
45130
#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
vector<pair<int, long long> > v[100004];
int parent[100004], level[100004];
long long dist[100004], size[100004], first[100004], f_[100004], g[100004],
    g_[100004];
int n;
int P[100004][30];
long long mu(long long x1, long long x2) { return (x1 * x2) % M; }
void preprocess(void) {
  int i, m;
  int j = n;
  for (i = 0; i <= j; i++)
    for (m = 0; (1 << m) <= j; m++) P[i][m] = -1;
  for (i = 1; i <= j; i++) P[i][0] = parent[i];
  for (m = 1; 1 << m <= j; m++)
    for (i = 1; i <= j; i++)
      if (P[i][m - 1] != -1) P[i][m] = P[P[i][m - 1]][m - 1];
}
int lca(int p, int q) {
  int tmp, log, i;
  if (level[p] < level[q]) tmp = p, p = q, q = tmp;
  for (log = 1; 1 << log <= level[p]; log++)
    ;
  log--;
  for (i = log; i >= 0; i--)
    if (level[p] - (1 << i) >= level[q]) p = P[p][i];
  if (p == q) return p;
  for (i = log; i >= 0; i--)
    if (P[p][i] != -1 && P[p][i] != P[q][i]) p = P[p][i], q = P[q][i];
  return parent[p];
}
void dfs(int node, int par) {
  size[node] = 1;
  for (int j = 0; j < v[node].size(); j++) {
    int ver = v[node][j].first;
    if (ver != par) {
      level[ver] = 1 + level[node];
      parent[ver] = node;
      dist[ver] = (dist[node] + v[node][j].second) % M;
      dfs(ver, node);
      size[node] += size[ver];
      first[node] =
          (first[node] + first[ver] + v[node][j].second * size[ver]) % M;
      g[node] = (g[node] + g[ver] + 2 * first[ver] * v[node][j].second +
                 mu(v[node][j].second, v[node][j].second) * size[ver]) %
                M;
    }
  }
}
void dfs1(int node, int par, long long d) {
  f_[node] =
      (f_[par] + first[par] - first[node] + d * (n - 2 * size[node])) % M;
  g_[node] = (g_[par] + g[par] - g[node] + mu(d, d) * (n - 4 * size[node]) +
              d * (2 * first[par] + 2 * f_[par] - 4 * first[node])) %
             M;
  for (int j = 0; j < v[node].size(); j++) {
    int ver = v[node][j].first;
    if (ver != par) dfs1(ver, node, v[node][j].second);
  }
}
long long ans(int u1, int v1) {
  int lc = lca(u1, v1);
  long long sum = 0;
  long long d = dist[u1] + dist[v1] - 2 * dist[lc];
  if (lc == v1) {
    sum = (g[u1] + g_[u1] - 2 * g_[v1] - 2 * mu(d, d) * (n - size[v1]) -
           4 * d * f_[v1]) %
          M;
  } else {
    sum = (2 * g[v1] - g_[u1] - g[u1] + 2 * mu(d, d) * size[v1] +
           4 * d * first[v1]) %
          M;
  }
  if (sum < 0) sum += M;
  return sum;
}
int main() {
  int a, b;
  long long c;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d", &a);
    scanf("%d", &b);
    scanf("%lld", &c);
    v[a].push_back(make_pair(b, c));
    v[b].push_back(make_pair(a, c));
  }
  parent[1] = 1;
  dist[1] = 0;
  level[1] = 1;
  dfs(1, 1);
  dfs1(1, 1, 0);
  preprocess();
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &a);
    scanf("%d", &b);
    cout << ans(a, b) << endl;
  }
}
