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>
const int mo = (int)1e9 + 7;
const int N = (int)1e5 + 5, logN = 17;
int n, q;
struct node;
struct edge {
  node *ed;
  edge *next;
  int w;
} E[N << 1], *newE = E;
struct data {
  long long sum2, sum, cnt;
  data operator*(long long t) {
    t %= mo;
    return (data){(sum2 + 2 * t * sum % mo + t * t % mo * cnt % mo) % mo,
                  (sum + t * cnt % mo) % mo, cnt};
  }
  void operator+=(const data &t) {
    (sum2 += t.sum2) %= mo;
    (sum += t.sum) %= mo;
    (cnt += t.cnt) %= mo;
  }
  void operator-=(const data &t) {
    (sum2 -= t.sum2) %= mo;
    (sum -= t.sum) %= mo;
    (cnt -= t.cnt) %= mo;
  }
};
struct node {
  edge *son;
  node *pa[logN];
  int dep;
  long long depth;
  data s, a;
  void dfs() {
    for (int j = 1; j < logN; j++) pa[j] = pa[j - 1]->pa[j - 1];
    s.cnt = 1;
    for (edge *e = son; e; e = e->next)
      if (e->ed != pa[0]) {
        e->ed->pa[0] = this;
        e->ed->dep = dep + 1;
        e->ed->depth = depth + e->w;
        e->ed->dfs();
        s += e->ed->s * e->w;
      }
  }
  void dfs2(data now) {
    now += s;
    a = now;
    for (edge *e = son; e; e = e->next)
      if (e->ed != pa[0]) {
        now -= e->ed->s * e->w;
        e->ed->dfs2(now * e->w);
        now += e->ed->s * e->w;
      }
  }
} V[N];
node *lca(node *u, node *v) {
  for (int j = logN - 1; j >= 0; j--)
    if (u->pa[j]->dep >= v->dep) u = u->pa[j];
  for (int j = logN - 1; j >= 0; j--)
    if (v->pa[j]->dep >= u->dep) v = v->pa[j];
  for (int j = logN - 1; j >= 0; j--)
    if (u->pa[j] != v->pa[j]) u = u->pa[j], v = v->pa[j];
  if (u == v) return u;
  return u->pa[0];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    *newE = (edge){V + y, V[x].son, w}, V[x].son = newE++;
    *newE = (edge){V + x, V[y].son, w}, V[y].son = newE++;
  }
  for (int j = 0; j < logN; j++) V[1].pa[j] = V + 1;
  V[1].dfs();
  V[1].dfs2((data){0, 0, 0});
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    node *w = lca(V + x, V + y);
    long long d = V[x].depth + V[y].depth - w->depth * 2;
    data a = V[y].s * d;
    if (w == V + y) {
      data c = V[y].a;
      c -= V[y].s;
      a = V[x].a;
      a -= c * d;
    }
    data b = V[x].a;
    b -= a;
    a -= b;
    printf("%I64d\n", (a.sum2 + mo) % mo);
  }
}
