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;
long long n, m, x, y, z, fa[300000], top[300000], dep[300000], sz[300000], tot,
    head[300000];
struct poly {
  long long f[3];
  poly() { memset(f, 0, sizeof f); }
  poly operator+(const poly &g) const {
    poly h;
    for (long long i = 0; i < 3; i++) h.f[i] = (f[i] + g.f[i]) % mod;
    return h;
  }
  void operator+=(const poly &g) { *this = *this + g; }
  poly operator-(const poly &g) const {
    poly h;
    for (long long i = 0; i < 3; i++) h.f[i] = (f[i] - g.f[i] + mod) % mod;
    return h;
  }
  void operator-=(const poly &g) { *this = *this - g; }
  long long operator()(long long k) const {
    long long ret = 0;
    k %= mod;
    for (long long i = 2; i >= 0; i--) ret = (ret * k + f[i]) % mod;
    return ret;
  }
  poly operator<<(long long k) const {
    poly g;
    k %= mod;
    for (long long i = 2; i >= 0; i--) {
      for (long long j = 2; j; j--) g.f[j] = (g.f[j] * k + g.f[j - 1]) % mod;
      g.f[0] = (g.f[0] * k + f[i]) % mod;
    }
    return g;
  }
} f[300000], g[300000];
struct edge {
  long long v, l, nxt;
} e[300000];
void add(long long x, long long y, long long z) {
  e[++tot].v = y;
  e[tot].l = z;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void dfs1(long long u) {
  f[u].f[2] = 1;
  for (long long i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa[u]) {
      fa[v] = u;
      dep[v] = dep[u] + e[i].l;
      dfs1(v);
      sz[u] += sz[v];
      f[u] += f[v] << e[i].l;
    }
  if (!sz[u]) ++sz[u];
}
void dfs2(long long u) {
  for (long long i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa[u])
      g[v] = g[u] + (f[u] << (-dep[u])) - (f[v] << (-dep[u] + e[i].l));
  if (!top[u]) top[u] = u;
  long long t = 0;
  for (long long i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa[u] && sz[v] > sz[t]) t = v;
  if (!t) return;
  top[t] = top[u];
  dfs2(t);
  for (long long i = head[u], v; i; i = e[i].nxt)
    if ((v = e[i].v) != fa[u] && v != t) dfs2(v);
}
long long lca(long long u, long long v) {
  for (; top[u] != top[v]; u = fa[top[u]])
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
  return dep[u] < dep[v] ? u : v;
}
long long ddd(long long x, long long y) {
  long long z = lca(x, y);
  if (z == y)
    return ((f[x](0) + g[x](dep[x]) - g[y](dep[x])) % mod + mod) % mod;
  return f[y](dep[x] + dep[y] - dep[z] * 2);
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i < n; i++) {
    scanf("%lld%lld%lld", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dfs1(1);
  dfs2(1);
  scanf("%lld", &m);
  for (long long i = 1; i <= m; i++) {
    scanf("%lld%lld", &x, &y);
    printf("%lld\n", ((ddd(x, y) * 2 - ddd(x, 1)) % mod + mod) % mod);
  }
  return 0;
}
