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;
int n, q;
vector<pair<int, int> > adj[100013];
int visited[100013];
int sizst[100013];
int sumst[100013];
int sqsst[100013];
int sumsq[100013];
int sumsm[100013];
int depth[100013];
int h[100013];
int ancestor[100013][20];
int up[100013];
int MOD = 1000000007;
void dfsLCA(int x) {
  visited[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    int next = adj[x][i].first;
    int dist = adj[x][i].second;
    if (!visited[next]) {
      h[next] = h[x] + 1;
      depth[next] = (depth[x] + dist) % MOD;
      dfsLCA(next);
      ancestor[next][0] = x;
      up[next] = dist;
    }
  }
  ancestor[x][0] = -1;
}
int findkth(int x, int k) {
  int where = x;
  for (int i = 19; i >= 0; i--)
    if (k & (1 << i)) where = ancestor[where][i];
  return where;
}
int lca(int a, int b) {
  if (h[a] > h[b]) swap(a, b);
  b = findkth(b, h[b] - h[a]);
  if (a == b) return a;
  for (int i = 19; i >= 0; i--) {
    if (h[a] < (1 << i)) continue;
    int na = ancestor[a][i];
    int nb = ancestor[b][i];
    if (na != nb) a = na, b = nb;
  }
  return ancestor[a][0];
}
void dfs(int x) {
  visited[x] = 1;
  sizst[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    int next = adj[x][i].first;
    if (visited[next]) continue;
    dfs(next);
    int dist = adj[x][i].second;
    sizst[x] += sizst[next];
    sumst[x] = (sumst[x] + sumst[next] + (long long)dist * sizst[next]) % MOD;
    int distsq = ((long long)dist * dist) % MOD;
    sqsst[x] = (sqsst[x] + (long long)distsq * sizst[next] +
                (long long)2 * dist * sumst[next] + sqsst[next]) %
               MOD;
  }
}
void dfs2(int x) {
  visited[x] = 1;
  if (x == 0) {
    sumsm[x] = sumst[x];
    sumsq[x] = sqsst[x];
  } else {
    int prev = ancestor[x][0];
    int dist = up[x];
    int distsq = ((long long)dist * dist) % MOD;
    int below = (sqsst[x] + (long long)2 * dist * sumst[x] +
                 (long long)sizst[x] * distsq) %
                MOD;
    int abovesq = (sumsq[prev] - below + MOD) % MOD;
    int abovesm =
        ((sumsm[prev] - sumst[x] - (long long)dist * sizst[x] + MOD) % MOD +
         MOD) %
        MOD;
    int num = n - sizst[x];
    sumsm[x] = (sumst[x] + abovesm + (long long)num * dist) % MOD;
    sumsq[x] = (sqsst[x] + (long long)distsq * num +
                ((long long)2 * dist * abovesm) % MOD + abovesq) %
               MOD;
  }
  for (int i = 0; i < adj[x].size(); i++)
    if (!visited[adj[x][i].first]) dfs2(adj[x][i].first);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    a -= 1;
    b -= 1;
    adj[a].push_back(make_pair(b, c));
    adj[b].push_back(make_pair(a, c));
  }
  dfsLCA(0);
  for (int i = 1; i < 20; i++) {
    for (int j = 0; j < n; j++) {
      if (h[j] < (1 << i))
        ancestor[j][i] = -1;
      else
        ancestor[j][i] = ancestor[ancestor[j][i - 1]][i - 1];
    }
  }
  for (int i = 0; i < n; i++) visited[i] = 0;
  dfs(0);
  for (int i = 0; i < n; i++) visited[i] = 0;
  dfs2(0);
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u -= 1;
    v -= 1;
    int l = lca(u, v);
    if (l == v) {
      int diff = (depth[u] - depth[v] + MOD) % MOD;
      int diffsq = ((long long)diff * diff) % MOD;
      int abovesq = (sumsq[v] - sqsst[v] + MOD) % MOD;
      int abovesm = (sumsm[v] - sumst[v] + MOD) % MOD;
      int tops = ((long long)diffsq * (n - sizst[v]) +
                  ((long long)2 * diff * abovesm) % MOD + abovesq) %
                 MOD;
      int part1 = (sumsq[u] - tops + MOD) % MOD;
      printf("%d\n", ((long long)2 * part1 - sumsq[u] + MOD) % MOD);
    } else {
      int diff =
          ((long long)depth[u] + depth[v] - 2 * depth[l] + 2 * MOD) % MOD;
      int diffsq = ((long long)diff * diff) % MOD;
      int part1 = ((long long)diffsq * sizst[v] +
                   ((long long)2 * diff * sumst[v]) % MOD + sqsst[v]) %
                  MOD;
      printf("%d\n", ((long long)2 * part1 - sumsq[u] + MOD) % MOD);
    }
  }
  return 0;
}
