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 N = 1e5 + 10;
const int S = 17;
vector<pair<int, int> > g[N];
int fa[N][S + 1], deep[N];
int dis[N];
int n, m;
struct node {
  int v2, v, sum;
} f[N], h[N];
const int mod = 1e9 + 7;
int add(int x, int y) {
  x += y;
  return x >= mod ? x - mod : x;
}
int sub(int x, int y) {
  x -= y;
  return x < 0 ? x + mod : x;
}
int mul(int x, int y) { return 1ll * x * y % mod; }
int calc(int x, int y) {
  int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x), y >>= 1;
  }
  return z;
}
node operator+(node a, node b) {
  a.v2 = add(a.v2, b.v2);
  a.v = add(a.v, b.v);
  a.sum = add(a.sum, b.sum);
  return a;
}
node operator-(node a, node b) {
  a.v2 = sub(a.v2, b.v2);
  a.v = sub(a.v, b.v);
  a.sum = sub(a.sum, b.sum);
  return a;
}
node link(node a, int x) {
  a.v2 = add(a.v2, add(mul(mul(2, a.v), x), mul(a.sum, mul(x, x))));
  a.v = add(a.v, mul(x, a.sum));
  return a;
}
void dfs(int x) {
  deep[x]++;
  for (int i = 0; fa[fa[x][i]][i]; i++) fa[x][i + 1] = fa[fa[x][i]][i];
  f[x].sum = 1;
  for (auto u : g[x])
    if (u.first != fa[x][0]) {
      fa[u.first][0] = x;
      deep[u.first] = deep[x];
      dis[u.first] = add(dis[x], u.second);
      dfs(u.first);
      f[x] = f[x] + link(f[u.first], u.second);
    }
}
int up(int x, int y) {
  for (int i = S; i >= 0; i--)
    if (y >= (1 << i)) y -= (1 << i), x = fa[x][i];
  return x;
}
int getlca(int x, int y) {
  if (deep[x] < deep[y]) swap(x, y);
  x = up(x, deep[x] - deep[y]);
  for (int i = S; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return x != y ? fa[x][0] : x;
}
void dfs(int x, node up) {
  h[x] = f[x] + up;
  for (auto u : g[x])
    if (u.first != fa[x][0]) {
      dfs(u.first, link(up + f[x] - link(f[u.first], u.second), u.second));
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    g[x].push_back({y, z});
    g[y].push_back({x, z});
  }
  dfs(1);
  dfs(1, node{0, 0, 0});
  scanf("%d", &m);
  while (m--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x == y) {
      int a = h[x].v2, b = f[x].v2;
      printf("%d\n", sub(mul(b, 2), a));
    } else {
      int lca = getlca(x, y);
      int v = sub(add(dis[x], dis[y]), mul(dis[lca], 2));
      if (lca == y) {
        int a = h[x].v2, b = link(h[y] - f[y], v).v2;
        printf("%d\n", sub(a, mul(b, 2)));
      } else {
        int a = h[x].v2, b = link(f[y], v).v2;
        printf("%d\n", sub(mul(b, 2), a));
      }
    }
  }
  return 0;
}
