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 md = 1000000007;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
}
inline int mul(int a, int b) { return (long long)a * b % md; }
const int N = 400010;
vector<pair<int, int> > g[N];
int cnt[N], sum[N], sumsq[N];
int tin[N], tout[N], TIME = 0;
int depth[N];
int pv[N];
const int LOG = 20;
int pr[N][LOG];
void dfs(int v, int pr) {
  tin[v] = ++TIME;
  int sz = g[v].size();
  cnt[v] = 1;
  sum[v] = 0;
  sumsq[v] = 0;
  for (int j = 0; j < sz; j++) {
    int u = g[v][j].first;
    if (u == pr) {
      continue;
    }
    pv[u] = v;
    int len = g[v][j].second;
    depth[u] = depth[v];
    add(depth[u], len);
    dfs(u, v);
    add(cnt[v], cnt[u]);
    add(sum[v], sum[u]);
    add(sum[v], mul(cnt[u], len));
    add(sumsq[v], sumsq[u]);
    add(sumsq[v], mul(cnt[u], mul(len, len)));
    add(sumsq[v], mul(mul(2, len), sum[u]));
  }
  tout[v] = ++TIME;
}
bool anc(int x, int y) { return (tin[x] <= tin[y] && tout[y] <= tout[x]); }
int lca(int x, int y) {
  if (anc(x, y)) return x;
  for (int j = LOG - 1; j >= 0; j--)
    if (!anc(pr[x][j], y)) x = pr[x][j];
  return pv[x];
}
int ups[N];
int up[N];
int n;
void get_up(int v, int pr) {
  int sz = g[v].size();
  for (int j = 0; j < sz; j++) {
    int u = g[v][j].first;
    if (u == pr) {
      continue;
    }
    int len = g[v][j].second;
    int no_s = sum[v];
    add(no_s, md - sum[u]);
    add(no_s, md - mul(cnt[u], len));
    add(no_s, ups[v]);
    int no_sq = sumsq[v];
    add(no_sq, md - sumsq[u]);
    add(no_sq, md - mul(cnt[u], mul(len, len)));
    add(no_sq, md - mul(mul(2, len), sum[u]));
    up[u] = up[v];
    add(up[u], no_sq);
    add(up[u], mul(n - cnt[u], mul(len, len)));
    add(up[u], mul(mul(2, len), no_s));
    ups[u] = ups[v];
    int no_sss = sum[v];
    add(no_sss, md - sum[u]);
    add(no_sss, md - mul(cnt[u], len));
    add(ups[u], no_sss);
    add(ups[u], mul(n - cnt[u], len));
    get_up(u, v);
  }
}
int res[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    g[i].clear();
  }
  for (int i = 0; i < n - 1; i++) {
    int foo, bar, len;
    scanf("%d %d %d", &foo, &bar, &len);
    foo--;
    bar--;
    g[foo].push_back(make_pair(bar, len));
    g[bar].push_back(make_pair(foo, len));
  }
  depth[0] = 0;
  dfs(0, -1);
  for (int i = 0; i < n; i++) pr[i][0] = pv[i];
  for (int j = 1; j < LOG; j++)
    for (int i = 0; i < n; i++) pr[i][j] = pr[pr[i][j - 1]][j - 1];
  ups[0] = 0;
  up[0] = 0;
  get_up(0, -1);
  for (int i = 0; i < n; i++) {
    res[i] = sumsq[i];
    add(res[i], up[i]);
  }
  int tt;
  scanf("%d", &tt);
  while (tt--) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    if (!anc(v, u)) {
      int len = depth[v];
      add(len, depth[u]);
      int z = lca(v, u);
      add(len, md - mul(2, depth[z]));
      int good = sumsq[v];
      add(good, mul(cnt[v], mul(len, len)));
      add(good, mul(mul(2, len), sum[v]));
      good = mul(good, 2);
      add(good, md - res[u]);
      printf("%d\n", good);
    } else {
      int len = depth[u];
      add(len, md - depth[v]);
      int good = up[v];
      add(good, mul(n - cnt[v], mul(len, len)));
      add(good, mul(mul(2, len), ups[v]));
      good = (res[u] - good + md) % md;
      good = mul(good, 2);
      add(good, md - res[u]);
      printf("%d\n", good);
    }
  }
  return 0;
}
