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 MAXL = 20, mod = 1e9 + 7;
int timer = 0, n, k, l, w, q, used[100005], ans, tin[100005], tout[100005];
long long sz[100005], dist[100005], distsq[100005], updist[100005],
    updistsq[100005], two[100005][20], twodist[100005][20];
vector<long long> d[100005];
vector<int> c[100005];
void dfs(int v, int p = 0, int w = 0) {
  tin[v] = timer++;
  sz[v] = 1;
  used[v] = 1;
  two[v][0] = p;
  twodist[v][0] = w;
  for (int i = 1; i < MAXL; i++) {
    twodist[v][i] = (twodist[v][i - 1] + twodist[two[v][i - 1]][i - 1]) % mod;
    two[v][i] = two[two[v][i - 1]][i - 1];
  }
  for (int i = 0; i < (int)c[v].size(); i++)
    if (used[c[v][i]] == 0) {
      dfs(c[v][i], v, d[v][i]);
      sz[v] += sz[c[v][i]];
      dist[v] = (dist[v] + dist[c[v][i]] + sz[c[v][i]] * d[v][i]) % mod;
      distsq[v] = (distsq[v] + distsq[c[v][i]] + 2 * d[v][i] * dist[c[v][i]] +
                   d[v][i] * ((d[v][i] * sz[c[v][i]]) % mod)) %
                  mod;
    }
  tout[v] = timer++;
}
void dfs2(int v) {
  used[v] = 1;
  for (int i = 0; i < (int)c[v].size(); i++)
    if (used[c[v][i]] == 0) {
      updist[c[v][i]] =
          ((updist[v] + dist[v] - (dist[c[v][i]] + sz[c[v][i]] * d[v][i]) +
            (n - sz[c[v][i]]) * d[v][i]) %
               mod +
           mod) %
          mod;
      long long l = ((updistsq[v] + distsq[v] -
                      (distsq[c[v][i]] + 2 * d[v][i] * dist[c[v][i]] +
                       d[v][i] * ((d[v][i] * sz[c[v][i]]) % mod))) %
                         mod +
                     mod) %
                    mod;
      updistsq[c[v][i]] =
          ((l - d[v][i] * ((d[v][i] * (n - sz[c[v][i]])) % mod) +
            2 * d[v][i] * updist[c[v][i]]) %
               mod +
           mod) %
          mod;
      dfs2(c[v][i]);
    }
}
bool upper(int v, int u) { return (tin[v] <= tin[u] && tout[v] >= tout[u]); }
int lca(int a, int b) {
  int res = 0;
  if (a == b) return 0;
  if (upper(a, b)) {
    for (int i = MAXL - 1; i >= 0; i--)
      if (!upper(two[b][i], a)) {
        res = (res + twodist[b][i]) % mod;
        b = two[b][i];
      }
    return (res + twodist[b][0]) % mod;
  }
  if (upper(b, a)) {
    for (int i = MAXL - 1; i >= 0; i--)
      if (!upper(two[a][i], b)) {
        res = (res + twodist[a][i]) % mod;
        a = two[a][i];
      }
    return (res + twodist[a][0]) % mod;
  }
  for (int i = MAXL - 1; i >= 0; i--)
    if (!upper(two[a][i], b)) {
      res = (res + twodist[a][i]) % mod;
      a = two[a][i];
    }
  for (int i = MAXL - 1; i >= 0; i--)
    if (!upper(two[b][i], a)) {
      res = (res + twodist[b][i]) % mod;
      b = two[b][i];
    }
  return ((res + twodist[a][0]) % mod + twodist[b][0]) % mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &k, &l, &w);
    k--;
    l--;
    c[k].push_back(l);
    c[l].push_back(k);
    d[k].push_back(w);
    d[l].push_back(w);
  }
  dfs(0);
  for (int i = 0; i < n; i++) used[i] = 0;
  dfs2(0);
  for (int i = 0; i < n; i++) used[i] = 0;
  scanf("%d", &q);
  for (int qq = 0; qq < q; qq++) {
    scanf("%d%d", &k, &l);
    k--;
    l--;
    long long x = lca(k, l);
    if (upper(l, k))
      ans = ((((distsq[k] + updistsq[k] -
                2 * (updistsq[l] + x * (x * (n - sz[l]) % mod) +
                     2 * x * updist[l])) %
               mod) +
              mod) %
             mod) %
            mod;
    else
      ans = ((2 * (x * ((x * sz[l]) % mod) + distsq[l] + 2 * x * dist[l]) -
              (distsq[k] + updistsq[k])) %
                 mod +
             mod) %
            mod;
    printf("%d\n", ans);
  }
  return 0;
}
