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 mod = 1000000007;
int modit(int x) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
int add(int a, int b) { return (a + b) % mod; }
int sub(int a, int b) { return add(a, mod - b); }
int mul(int a, int b) { return (long long)a * b % mod; }
int sq(int x) { return mul(x, x); }
struct PowerSums {
  int sum[3];
  PowerSums() { memset(sum, 0, sizeof sum); }
  PowerSums append(int x) {
    PowerSums ret;
    ret.sum[0] = sum[0];
    x = modit(x);
    ret.sum[2] = add(sum[2], add(mul(2, mul(x, sum[1])), mul(sq(x), sum[0])));
    ret.sum[1] = add(sum[1], mul(x, sum[0]));
    return ret;
  }
  PowerSums add_zero() {
    PowerSums ret = *this;
    ret.sum[0] = add(1, sum[0]);
    return ret;
  }
  int get() { return sum[2]; }
};
PowerSums apply(const PowerSums &a, const PowerSums &b, int f(int, int)) {
  PowerSums ret;
  for (int i = (0); i < (3); ++i) ret.sum[i] = f(a.sum[i], b.sum[i]);
  return ret;
}
PowerSums operator+(const PowerSums &a, const PowerSums &b) {
  return apply(a, b, add);
}
PowerSums operator-(const PowerSums &a, const PowerSums &b) {
  return apply(a, b, sub);
}
const int MAXN = 100005;
const int MAXL = 18;
PowerSums down[MAXN];
PowerSums tot[MAXN];
PowerSums up[MAXN];
int N, Q;
vector<pair<int, int> > T[MAXN];
int rdist[MAXN];
void calc_down(int i, int dad, int rd) {
  rdist[i] = rd;
  for (auto e : T[i]) {
    int j = e.first;
    int d = e.second;
    if (j != dad) {
      calc_down(j, i, add(rd, d));
      down[i] = down[i] + down[j].append(d);
    }
  }
  down[i] = down[i].add_zero();
}
void calc_tot(int i, int dad) {
  for (auto e : T[i]) {
    int j = e.first;
    int d = e.second;
    if (j != dad) {
      tot[j] = down[j] + (tot[i] - down[j].append(d)).append(d);
      calc_tot(j, i);
    }
  }
  up[i] = tot[i] - down[i];
}
void calc_all() {
  calc_down(0, -1, 0);
  tot[0] = down[0];
  calc_tot(0, -1);
}
int dad[MAXN];
int dep[MAXN];
int anc[MAXN][MAXL];
void dfs(int i, int d) {
  dep[i] = d;
  for (auto it : T[i]) {
    int j = it.first;
    if (j != dad[i]) {
      dad[j] = i;
      dfs(j, d + 1);
    }
  }
}
void lca_precalc() {
  dfs(0, 0);
  for (int i = (0); i < (N); ++i) anc[i][0] = dad[i];
  for (int p = (1); p < (MAXL); ++p)
    for (int i = (0); i < (N); ++i) anc[i][p] = anc[anc[i][p - 1]][p - 1];
}
int lca(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  for (int p = MAXL - 1; p >= 0; --p)
    if (dep[anc[a][p]] >= dep[b]) a = anc[a][p];
  if (a == b) return a;
  for (int p = MAXL - 1; p >= 0; --p)
    if (anc[a][p] != anc[b][p]) a = anc[a][p], b = anc[b][p];
  return dad[a];
}
int dist(int a, int b) {
  return sub(add(rdist[a], rdist[b]), mul(2, rdist[lca(a, b)]));
}
bool is_ancestor(int a, int b) { return lca(a, b) == a; }
int main(void) {
  scanf("%d", &N);
  for (int i = (0); i < (N - 1); ++i) {
    int a, b, d;
    scanf("%d%d%d", &a, &b, &d);
    --a;
    --b;
    T[a].push_back(pair<int, int>(b, d));
    T[b].push_back(pair<int, int>(a, d));
  }
  lca_precalc();
  calc_all();
  scanf("%d", &Q);
  for (int i = (0); i < (Q); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    if (is_ancestor(v, u))
      printf("%d\n", sub(tot[u].get(), mul(2, up[v].append(dist(u, v)).get())));
    else
      printf("%d\n",
             sub(mul(2, down[v].append(dist(u, v)).get()), tot[u].get()));
  }
  return 0;
}
