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 long long MOD = 1000000007;
const int MAXN = 100010;
int n;
struct edge {
  int t;
  long long w;
  edge *next;
} epool[MAXN * 2], *e[MAXN];
int psz;
int dep[MAXN], fa[MAXN][21];
long long dis[MAXN];
struct S {
  long long size, sum, sum2;
  S() {}
  S(long long size, long long sum, long long sum2)
      : size(size), sum(sum), sum2(sum2) {}
  S operator+(long long d) const {
    return S(size, (sum + d * size % MOD) % MOD,
             (sum2 + 2 * d * sum % MOD + d * d % MOD * size % MOD) % MOD);
  }
  S &operator+=(const S &b) {
    size += b.size, sum += b.sum, sum2 += b.sum2;
    size %= MOD, sum %= MOD, sum2 %= MOD;
    return *this;
  }
  S &operator-=(const S &b) {
    size += MOD - b.size, sum += MOD - b.sum, sum2 += MOD - b.sum2;
    size %= MOD, sum %= MOD, sum2 %= MOD;
    return *this;
  }
} f[MAXN], g[MAXN];
void addedge(int u, int v, long long w) {
  epool[psz] = (edge){v, w, e[u]};
  e[u] = epool + psz++;
}
void init() {
  psz = 0, memset(e, 0, sizeof(e));
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    addedge(u, v, w % MOD);
    addedge(v, u, w % MOD);
  }
}
void dfs1(int u) {
  f[u] = S(1, 0, 0);
  for (int i = 0; i < 20; ++i) fa[u][i + 1] = fa[fa[u][i]][i];
  for (edge *i = e[u]; i; i = i->next) {
    int v = i->t;
    if (v == fa[u][0]) continue;
    fa[v][0] = u;
    dep[v] = dep[u] + 1;
    dis[v] = (dis[u] + i->w) % MOD;
    dfs1(v);
    f[u] += f[v] + i->w;
  }
}
void dfs2(int u) {
  for (edge *i = e[u]; i; i = i->next) {
    int v = i->t;
    if (v == fa[u][0]) continue;
    g[v] = g[u] + i->w;
    g[v] += f[u] + i->w;
    g[v] -= f[v] + i->w * 2 % MOD;
    dfs2(v);
  }
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 20; i >= 0; --i) {
    if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; --i) {
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  }
  return fa[u][0];
}
void solve() {
  int q;
  scanf("%d", &q);
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    int w = lca(u, v);
    long long d = ((dis[u] + dis[v] - dis[w] * 2) % MOD + MOD) % MOD;
    S s1, s2;
    if (w == v) {
      s2 = g[v] + d;
      s1 = f[u];
      s1 += g[u];
      s1 -= s2;
    } else {
      s1 = f[v] + d;
      s2 = f[u];
      s2 += g[u];
      s2 -= s1;
    }
    s1 -= s2;
    printf("%d\n", (int)s1.sum2);
  }
}
int main() {
  init();
  dep[1] = 1, dis[1] = 0;
  dfs1(1);
  dfs2(1);
  solve();
}
