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;
long long const M = 1e5 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
long long par[M][30], dp[M], zt[M], sum[M], val[M][30], sum2[M], n, dp2[M],
    len = 0, h[M];
vector<long long> adj[M], edg[M], inc[M];
long long mrg(long long x, long long y, long long sm, long long cnt) {
  long long ans = (x + ((y * y) % mod) * cnt + ((sm * y) % mod) * 2) % mod;
  return ans;
}
void dfs_down(long long v, long long l, long long c) {
  zt[v] = 1;
  val[v][0] = c;
  h[v] = h[l] + 1;
  par[v][0] = l;
  for (long long i = 1; i <= 20; i++) {
    par[v][i] = par[par[v][i - 1]][i - 1];
    val[v][i] = val[par[v][i - 1]][i - 1] + val[v][i - 1];
    val[v][i] %= mod;
  }
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u != l) dfs_down(u, v, edg[v][i]);
    if (u == l) {
      inc[v].push_back(0);
      continue;
    }
    zt[v] += zt[u];
    sum[v] += sum[u];
    sum[v] %= mod;
    sum[v] += edg[v][i] * zt[u];
    sum[v] %= mod;
    long long res = mrg(dp[u], edg[v][i], sum[u], zt[u]);
    inc[v].push_back(res);
    dp[v] += res;
    dp[v] %= mod;
  }
  dp2[v] = dp[v];
}
void dfs_up(long long v, long long l, long long c, long long cc) {
  if (v == 1) sum2[v] = sum[v];
  if (v != 1) {
    sum2[v] += (sum2[l] - ((cc * zt[v] + sum[v]) % mod) + mod) % mod;
    sum2[v] %= mod;
    long long res = dp[l] - c;
    res += mod;
    res %= mod;
    dp[v] += mrg(res, cc, sum2[v], n - zt[v]);
    dp[v] %= mod;
    sum2[v] += sum[v];
    sum2[v] %= mod;
    sum2[v] += cc * (n - zt[v]);
    sum2[v] %= mod;
  }
  for (long long i = 0; i < adj[v].size(); i++) {
    long long u = adj[v][i];
    if (u != l) dfs_up(u, v, inc[v][i], edg[v][i]);
  }
}
long long lca(long long v, long long u) {
  if (h[v] < h[u]) swap(v, u);
  for (long long i = 20; i >= 0; i--) {
    if (h[v] - (1 << i) >= h[u]) len += val[v][i], v = par[v][i];
  }
  len %= mod;
  if (v == u) return v;
  for (long long i = 20; i >= 0; i--) {
    if (par[v][i] != par[u][i])
      len += val[v][i], len += val[u][i], v = par[v][i], u = par[u][i];
  }
  len += val[v][0];
  len += val[u][0];
  len %= mod;
  v = par[v][0];
  return v;
}
int32_t main() {
  cin >> n;
  for (long long i = 1; i <= n - 1; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    adj[a].push_back(b);
    adj[b].push_back(a);
    edg[a].push_back(c);
    edg[b].push_back(c);
  }
  dfs_down(1, 0, 0);
  dfs_up(1, 0, 0, 0);
  long long q;
  cin >> q;
  while (q--) {
    long long v, u;
    cin >> u >> v;
    len = 0;
    long long lc = lca(u, v);
    len %= mod;
    long long all = 0;
    if (lc != v) {
      all = -dp[u];
      all += mod;
      all %= mod;
      long long res = mrg(dp2[v], len, sum[v], zt[v]);
      all += res * 2;
      all %= mod;
    } else {
      all = dp[u];
      long long res = sum2[v] - sum[v];
      long long bad = dp[v] - dp2[v];
      bad += mod;
      bad %= mod;
      res += mod;
      res %= mod;
      long long hlp = mrg(bad, len, res, n - zt[v]);
      all -= 2 * hlp;
      all += 3 * mod;
      all %= mod;
    }
    cout << all << "\n";
  }
}
