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 MX = 100005, LG = 18, mod = 1000000007;
int n, p[MX][LG], lvl[MX], q;
long long d[MX], d2[MX], sz[MX], sd[MX], sd2[MX], tor[MX];
vector<pair<int, long long> > adj[MX];
int LCA(int u, int v) {
  if (lvl[u] < lvl[v]) swap(u, v);
  for (int j = 0; j < LG; j++)
    if (lvl[u] - lvl[v] & (1 << j)) u = p[u][j];
  if (u == v) return u;
  for (int j = LG - 1; j >= 0; j--)
    if (p[u][j] != p[v][j]) {
      u = p[u][j];
      v = p[v][j];
    }
  return p[u][0];
}
void pre(int u) {
  sz[u] = 1;
  for (pair<int, long long> &v : adj[u])
    if (v.first != p[u][0]) {
      lvl[v.first] = lvl[u] + 1;
      p[v.first][0] = u;
      tor[v.first] = (tor[u] + v.second) % mod;
      pre(v.first);
      sz[u] += sz[v.first];
      (sd[u] += (sd[v.first] + sz[v.first] * v.second)) %= mod;
      (sd2[u] += (sd2[v.first] + 2 * sd[v.first] * v.second +
                  v.second * v.second % mod * sz[v.first])) %= mod;
    }
}
void dfs(int u) {
  for (pair<int, long long> &v : adj[u])
    if (v.first != p[u][0]) {
      d[v.first] = (d[u] + v.second * (n - 2 * sz[v.first])) % mod;
      d2[v.first] =
          (d2[u] +
           2 * v.second *
               (d[u] - 2 * (sd[v.first] + v.second * sz[v.first]) % mod) % mod +
           v.second * v.second % mod * n) %
          mod;
      dfs(v.first);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    adj[a].emplace_back(b, c);
    adj[b].emplace_back(a, c);
  }
  p[1][0] = 1;
  pre(1);
  for (int j = 1; j < LG; j++)
    for (int i = 1; i <= n; i++) p[i][j] = p[p[i][j - 1]][j - 1];
  d[1] = sd[1];
  d2[1] = sd2[1];
  dfs(1);
  cin >> q;
  while (q--) {
    int u, v;
    cin >> u >> v;
    int lca = LCA(u, v);
    long long res = lca == v ? d2[u] : -d2[u];
    long long dis = (tor[u] + tor[v] - 2 * tor[lca]) % mod;
    if (lca == v) {
      (res -= 2 * (((d2[v] - sd2[v]) + 2 * (d[v] - sd[v]) * dis +
                    dis * dis % mod * (n - sz[v])) %
                   mod)) %= mod;
    } else {
      (res +=
       2 * ((sd2[v] + 2 * sd[v] * dis + dis * dis % mod * sz[v]) % mod)) %= mod;
    }
    cout << ((((res) % (mod)) + (mod)) % (mod)) << '\n';
  }
  return 0;
}
