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:

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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 int mod = 1e9 + 7;
struct sol_t {
  int many;
  int sum2;
  int sum1;
  sol_t() { many = sum1 = sum2 = 0; }
  void include(const sol_t &other) {
    many = (many + other.many) % mod;
    sum2 = (sum2 + other.sum2) % mod;
    sum1 = (sum1 + other.sum1) % mod;
  }
  void uninclude(const sol_t &other) {
    many = (many - other.many + mod) % mod;
    sum2 = (sum2 - other.sum2 + mod) % mod;
    sum1 = (sum1 - other.sum1 + mod) % mod;
  }
  void increase_all(int by) {
    sum2 =
        (sum2 + (long long)many * by % mod * by % mod + 2LL * by * sum1) % mod;
    sum1 = (sum1 + (long long)many * by) % mod;
  }
};
vector<pair<int, int> > adj[100000];
int height[100000];
long long depth[100000];
sol_t down[100000];
sol_t up[100000];
sol_t full[100000];
vector<int> path;
vector<long long> pathsum;
vector<pair<int, int> > queries[100000];
sol_t result[100000];
int logn;
int parent[100000][16 + 1];
int goup(int a, int dist) {
  for (int l = 0; (1 << l) <= dist; l++)
    if (dist & (1 << l)) a = parent[a][l];
  return a;
}
int lca(int a, int b) {
  if (height[a] < height[b]) swap(a, b);
  a = goup(a, height[a] - height[b]);
  if (a == b) return a;
  for (int l = logn; l >= 0; l--)
    if (parent[a][l] != parent[b][l]) {
      a = parent[a][l];
      b = parent[b][l];
    }
  return parent[a][0];
}
long long dist(int a, int b) {
  return depth[a] + depth[b] - 2 * depth[lca(a, b)];
}
void dfs3(int u, int p, long long sum = 0) {
  path.push_back(u);
  pathsum.push_back(sum);
  for (pair<int, int> query : queries[u]) {
    int v = query.first;
    int qid = query.second;
    if (height[v] < (int)path.size() && path[height[v]] == v) {
      result[qid].include(full[u]);
      sol_t temp = up[v];
      temp.increase_all((sum - pathsum[height[v]]) % mod);
      result[qid].uninclude(temp);
      result[qid].uninclude(temp);
    } else {
      result[qid].uninclude(full[u]);
      sol_t temp = down[v];
      temp.increase_all(dist(u, v) % mod);
      result[qid].include(temp);
      result[qid].include(temp);
    }
  }
  for (const pair<int, int> &pp : adj[u]) {
    int v2 = pp.first, dist = pp.second;
    if (v2 == p) continue;
    dfs3(v2, u, sum + dist);
  }
  path.pop_back();
  pathsum.pop_back();
}
void dfs2(int v, int p, sol_t got) {
  up[v] = got;
  full[v] = up[v];
  full[v].include(down[v]);
  vector<pair<int, int> > child;
  for (const pair<int, int> &pp : adj[v]) {
    int v2 = pp.first, dist = pp.second;
    if (v2 == p) continue;
    child.push_back(make_pair(v2, dist));
  }
  int nc = child.size();
  sol_t sum = down[v];
  for (int c = 0; c < nc; c++) {
    sol_t just = sum;
    {
      sol_t temp = down[child[c].first];
      temp.increase_all(child[c].second);
      just.uninclude(temp);
    }
    just.include(got);
    just.increase_all(child[c].second);
    dfs2(child[c].first, v, just);
  }
}
void dfs(int v, int p, int h, long long d) {
  for (int l = 1; l <= logn; l++)
    parent[v][l] = parent[parent[v][l - 1]][l - 1];
  height[v] = h;
  depth[v] = d;
  down[v] = sol_t();
  down[v].many = 1;
  for (const pair<int, int> &pp : adj[v]) {
    int v2 = pp.first, dist = pp.second;
    if (v2 == p) continue;
    parent[v2][0] = v;
    dfs(v2, v, h + 1, d + dist);
    sol_t temp = down[v2];
    temp.increase_all(dist);
    down[v].include(temp);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  logn = 0;
  while ((1 << (logn + 1)) <= n) logn++;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--, b--;
    adj[a].push_back(make_pair(b, c));
    adj[b].push_back(make_pair(a, c));
  }
  parent[0][0] = 0;
  dfs(0, -1, 0, 0);
  sol_t temp;
  dfs2(0, -1, temp);
  int nq;
  scanf("%d", &nq);
  for (int q = 0; q < nq; q++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--, v--;
    queries[u].push_back(make_pair(v, q));
  }
  dfs3(0, -1);
  for (int q = 0; q < nq; q++) printf("%d\n", result[q].sum2);
  return 0;
}
