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;
int n;
vector<pair<int, int> > edge[100005];
long long sum[100005];
long long sum2[100005];
long long up[100005];
long long up2[100005];
int s[100005];
int lca[100005][20];
int dep[100005];
long long dist[100005];
void dfs(int v, int u) {
  s[v] = 1;
  lca[v][0] = u;
  for (int i = 0; i < edge[v].size(); i++) {
    if (edge[v][i].first == u) continue;
    int x = edge[v][i].first;
    dfs(x, v);
    sum[v] += sum[x] + 1LL * s[x] * edge[v][i].second;
    sum2[v] += sum2[x] +
               1LL * s[x] * edge[v][i].second % 1000000007 * edge[v][i].second +
               2LL * edge[v][i].second * sum[x];
    sum[v] %= 1000000007;
    sum2[v] %= 1000000007;
    s[v] += s[x];
  }
}
void dfs2(int v, int u, int c) {
  if (u != -1) {
    dep[v] = dep[u] + 1;
    dist[v] = dist[u] + c;
    long long x = sum2[v] + 2LL * c * sum[v] + 1LL * c * c % 1000000007 * s[v];
    x %= 1000000007;
    long long y = sum[v] + 1LL * c * s[v];
    y %= 1000000007;
    long long q = (sum2[u] + up2[u] - x + 1000000007) % 1000000007;
    long long r = (sum[u] + up[u] - y + 1000000007) % 1000000007;
    up[v] = (r + 1LL * (n - s[v]) * c) % 1000000007;
    up2[v] =
        (q + 2LL * r * c + 1LL * c * c % 1000000007 * (n - s[v])) % 1000000007;
  }
  for (int i = 0; i < edge[v].size(); i++)
    if (u != edge[v][i].first) dfs2(edge[v][i].first, v, edge[v][i].second);
}
int calc(int a, int b) {
  if (dep[a] > dep[b]) swap(a, b);
  for (int i = 0; i < 20; i++) {
    if ((((dep[b] - dep[a]) >> i) & 1)) {
      b = lca[b][i];
    }
  }
  if (a == b) return a;
  for (int i = 19; i >= 0; i--) {
    if (lca[a][i] != lca[b][i]) {
      a = lca[a][i];
      b = lca[b][i];
    }
  }
  return lca[a][0];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    edge[a].push_back(make_pair(b, c));
    edge[b].push_back(make_pair(a, c));
  }
  dfs(1, -1);
  dfs2(1, -1, -1);
  int q;
  scanf("%d", &q);
  for (int i = 0; i < 19; i++)
    for (int j = 1; j <= n; j++)
      if (lca[j][i] == -1)
        lca[j][i + 1] = -1;
      else
        lca[j][i + 1] = lca[lca[j][i]][i];
  for (int i = 0; i < q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    int x = calc(u, v);
    long long d = dist[u] + dist[v] - dist[x] * 2;
    d %= 1000000007;
    if (x == v && u != v) {
      long long a =
          up2[v] + 2LL * d * up[v] + 1LL * d * d % 1000000007 * (n - s[v]);
      a %= 1000000007;
      printf("%lld\n",
             (sum2[u] + up2[u] - 2 * a + 1000000007 + 1000000007) % 1000000007);
    } else {
      long long a =
          sum2[v] + 2LL * d * sum[v] + 1LL * d * d % 1000000007 * (s[v]);
      a %= 1000000007;
      printf("%lld\n",
             (2 * a - sum2[u] - up2[u] + 1000000007 + 1000000007) % 1000000007);
    }
  }
}
