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 mo = int(1e9 + 7);
struct edge {
  int s, t, d, n;
} e[200010];
int n, q, h[100010], f[100010][6], g[100010][20], gg[100010][20], dp[100010];
int sqr_mod(int x) { return 1LL * x * x % mo; }
void dfs(int x, int fa) {
  f[x][3] = 1;
  for (int i = h[x], y, d; y = e[i].t, d = e[i].d, i; i = e[i].n)
    if (y != fa) {
      dfs(y, x);
      f[x][1] =
          (f[x][1] + f[y][1] + 2LL * f[y][2] * d + 1LL * sqr_mod(d) * f[y][3]) %
          mo;
      f[x][2] = (f[x][2] + f[y][2] + 1LL * d * f[y][3]) % mo;
      f[x][3] += f[y][3];
    }
}
void dfs(int x, int fa, int d) {
  if (fa) {
    f[x][4] = ((1LL * (n - 4 * f[x][3]) * sqr_mod(d) +
                (2LL * f[fa][5] + 2LL * f[fa][2] - 4LL * f[x][2]) * d +
                f[fa][4] + f[fa][1] - f[x][1]) %
                   mo +
               mo) %
              mo;
    f[x][5] =
        ((1LL * (n - 2 * f[x][3]) * d + f[fa][5] + f[fa][2] - f[x][2]) % mo +
         mo) %
        mo;
  }
  g[x][0] = fa, gg[x][0] = d, dp[x] = dp[fa] + 1;
  for (int i = h[x]; i; i = e[i].n)
    if (e[i].t != fa) dfs(e[i].t, x, e[i].d);
}
void LCA(int x, int y, int &t, int &d) {
  if (dp[x] > dp[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (dp[x] <= dp[g[y][i]]) d = (d + gg[y][i]) % mo, y = g[y][i];
  if (x == y) return void(t = x);
  for (int i = 19; i >= 0; i--)
    if (g[x][i] != g[y][i])
      d = ((d + gg[x][i]) % mo + gg[y][i]) % mo, x = g[x][i], y = g[y][i];
  d = ((d + gg[x][0]) % mo + gg[y][0]) % mo, t = g[x][0];
}
void work() {
  scanf("%d", &n);
  for (int i = 1, s, t, d, tot = 0; i < n; i++) {
    scanf("%d %d %d", &s, &t, &d);
    e[++tot] = (edge){s, t, d, h[s]}, h[s] = tot;
    e[++tot] = (edge){t, s, d, h[t]}, h[t] = tot;
  }
  dfs(1, 0), dfs(1, 0, 0);
  for (int i = 1; i <= n; i++) f[i][0] = (f[i][1] + f[i][4]) % mo;
  for (int j = 1; j < 20; j++)
    for (int i = 1; i <= n; i++)
      g[i][j] = g[g[i][j - 1]][j - 1],
      gg[i][j] = (gg[i][j - 1] + gg[g[i][j - 1]][j - 1]) % mo;
  scanf("%d", &q);
  for (int i = 1, x, y, t, d, ans; i <= q; i++) {
    scanf("%d %d", &x, &y), d = 0;
    LCA(x, y, t, d);
    if (t == y)
      ans = (((f[x][1] + f[x][4] - f[y][4] - 2LL * d * f[y][5] -
               1LL * (n - f[y][3]) * sqr_mod(d)) %
                  mo * 2LL -
              f[x][0]) %
                 mo +
             mo) %
            mo;
    else
      ans = (((f[y][1] + 2LL * d * f[y][2] + 1LL * f[y][3] * sqr_mod(d)) % mo *
                  2LL -
              f[x][0]) %
                 mo +
             mo) %
            mo;
    printf("%d\n", ans);
  }
}
int main() {
  work();
  return 0;
}
