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 maxn = 1e5 + 10;
const int mod = 1e9 + 7;
int n;
long long h[maxn], answer[maxn];
vector<pair<int, int> > t[maxn], query[maxn];
int Time = 0, st[maxn], ft[maxn];
struct node {
  long long d1;
  long long d2;
  long long lazy;
  node() { d1 = 0, d2 = 0, lazy = 0; }
};
node seg[4 * maxn];
void propagate(int, int, int);
int get(int id, int L, int R, int l, int r) {
  if (r <= L or R <= l) return 0;
  if (l <= L and R <= r) return seg[id].d2;
  propagate(id, L, R);
  int mid = (L + R) >> 1;
  return (get(2 * id + 0, L, mid, l, r) + get(2 * id + 1, mid, R, l, r)) % mod;
}
void add(int id, int L, int R, int l, int r, long long val) {
  if (r <= L or R <= l) return;
  if (l <= L and R <= r) {
    val %= mod;
    val = (val + mod) % mod;
    seg[id].d2 = (seg[id].d2 + 1ll * (R - L) * val % mod * val % mod +
                  2ll * val % mod * seg[id].d1 % mod) %
                 mod;
    seg[id].d1 = (seg[id].d1 + 1ll * (R - L) * val % mod) % mod;
    seg[id].lazy = (seg[id].lazy + val) % mod;
    return;
  }
  propagate(id, L, R);
  int mid = (L + R) >> 1;
  add(2 * id + 0, L, mid, l, r, val);
  add(2 * id + 1, mid, R, l, r, val);
  seg[id].d1 = (seg[2 * id + 0].d1 + seg[2 * id + 1].d1) % mod;
  seg[id].d2 = (seg[2 * id + 0].d2 + seg[2 * id + 1].d2) % mod;
}
void propagate(int id, int L, int R) {
  if (seg[id].lazy == 0) return;
  int mid = (L + R) >> 1;
  add(2 * id + 0, L, mid, L, mid, seg[id].lazy);
  add(2 * id + 1, mid, R, mid, R, seg[id].lazy);
  seg[id].lazy = 0;
}
void dfs(int v, int par = -1) {
  for (auto Q : query[v]) {
    int idx = Q.first, u = Q.second;
    answer[idx] = 2ll * get(1, 0, n, st[u], ft[u]) % mod - get(1, 0, n, 0, n);
    answer[idx] = (answer[idx] + mod) % mod;
  }
  for (auto edge : t[v]) {
    int u = edge.first, w = edge.second;
    if (u != par) {
      add(1, 0, n, 0, n, +w);
      add(1, 0, n, st[u], ft[u], -2 * w);
      dfs(u, v);
      add(1, 0, n, st[u], ft[u], +2 * w);
      add(1, 0, n, 0, n, -w);
    }
  }
}
void dfs_init(int v, int par = -1) {
  st[v] = Time++;
  for (auto edge : t[v]) {
    int u = edge.first, w = edge.second;
    if (u != par) {
      h[u] = h[v] + w;
      dfs_init(u, v);
    }
  }
  ft[v] = Time;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v--, u--;
    t[v].push_back({u, w});
    t[u].push_back({v, w});
  }
  dfs_init(0);
  for (int i = 0; i < n; i++) add(1, 0, n, st[i], st[i] + 1, h[i]);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, u;
    cin >> v >> u;
    v--, u--;
    query[v].push_back({i, u});
  }
  dfs(0);
  for (int i = 0; i < q; i++) cout << answer[i] << endl;
}
