#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100005;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
vector<pair<int, int> > g[Maxn];
int par[Maxn][20], dep[Maxn], n, m, q;
long long dist[Maxn][50], sum[Maxn];
vector<int> qf;
bool vis[Maxn], low[Maxn];
void tarjan(int u, int pre) {
  vis[u] = true;
  for (register int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    if (v ^ pre) {
      if (!vis[v]) {
        par[v][0] = u;
        dep[v] = dep[u] + 1;
        sum[v] = sum[u] + g[u][i].second;
        tarjan(v, u);
      } else
        low[u] = low[v] = true;
    }
  }
}
int lca(int u, int v) {
  if (dep[u] > dep[v]) swap(u, v);
  int d = dep[v] - dep[u];
  for (register int i = 0; i < 20; i++)
    if (d >> i & 1) v = par[v][i];
  if (u == v) return v;
  for (register int i = 19; i >= 0; i--) {
    if (par[u][i] ^ par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  }
  return par[u][0];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, x, y, cost; i < m; i++) {
    scanf("%d%d%d", &x, &y, &cost);
    g[x].push_back(make_pair(y, cost));
    g[y].push_back(make_pair(x, cost));
  }
  par[1][0] = 1;
  tarjan(1, 1);
  for (register int j = 1; j < 20; j++)
    for (register int i = 1; i <= n; i++) par[i][j] = par[par[i][j - 1]][j - 1];
  for (register int i = 1; i <= n; i++)
    if (low[i]) qf.push_back(i);
  for (register int num = 0; num < qf.size(); num++) {
    for (register int i = 1; i <= n; i++) dist[i][num] = lnf;
    priority_queue<pair<long long, int> > pq;
    dist[qf[num]][num] = 0;
    pq.push(make_pair(0, qf[num]));
    while (!pq.empty()) {
      pair<long long, int> cur = pq.top();
      pq.pop();
      if (dist[cur.second][num] ^ -cur.first) continue;
      int u = cur.second;
      for (register int i = 0; i < g[u].size(); i++) {
        int v = g[u][i].first;
        if (dist[v][num] > dist[u][num] + g[u][i].second) {
          dist[v][num] = dist[u][num] + g[u][i].second;
          pq.push(make_pair(-dist[v][num], v));
        }
      }
    }
  }
  scanf("%d", &q);
  for (int u, v, lc; q--;) {
    scanf("%d%d", &u, &v);
    lc = lca(u, v);
    long long ans = sum[u] + sum[v] - 2 * sum[lc];
    for (int i = 0; i < qf.size(); i++) ans = min(ans, dist[u][i] + dist[v][i]);
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given a weighed undirected connected graph, consisting of n vertices and m edges.

You should answer q queries, the i-th query is to find the shortest distance between vertices u_i and v_i.

Input

The first line contains two integers n and m~(1 ≤ n, m ≤ 10^5, m - n ≤ 20) — the number of vertices and edges in the graph.

Next m lines contain the edges: the i-th edge is a triple of integers v_i, u_i, d_i~(1 ≤ u_i, v_i ≤ n, 1 ≤ d_i ≤ 10^9, u_i ≠ v_i). This triple means that there is an edge between vertices u_i and v_i of weight d_i. It is guaranteed that graph contains no self-loops and multiple edges.

The next line contains a single integer q~(1 ≤ q ≤ 10^5) — the number of queries.

Each of the next q lines contains two integers u_i and v_i~(1 ≤ u_i, v_i ≤ n) — descriptions of the queries.

Pay attention to the restriction m - n ~ ≤ ~ 20.

Output

Print q lines.

The i-th line should contain the answer to the i-th query — the shortest distance between vertices u_i and v_i.

Examples

Input

3 3
1 2 3
2 3 1
3 1 5
3
1 2
1 3
2 3


Output

3
4
1


Input

8 13
1 2 4
2 3 6
3 4 1
4 5 12
5 6 3
6 7 8
7 8 7
1 4 1
1 8 3
2 6 9
2 7 1
4 6 3
6 8 2
8
1 5
1 7
2 3
2 8
3 7
3 4
6 8
7 8


Output

7
5
6
7
7
1
2
7

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100005;
const int inf = 0x3f3f3f3f;
const long long lnf = 0x3f3f3f3f3f3f3f3f;
vector<pair<int, int> > g[Maxn];
int par[Maxn][20], dep[Maxn], n, m, q;
long long dist[Maxn][50], sum[Maxn];
vector<int> qf;
bool vis[Maxn], low[Maxn];
void tarjan(int u, int pre) {
  vis[u] = true;
  for (register int i = 0; i < g[u].size(); i++) {
    int v = g[u][i].first;
    if (v ^ pre) {
      if (!vis[v]) {
        par[v][0] = u;
        dep[v] = dep[u] + 1;
        sum[v] = sum[u] + g[u][i].second;
        tarjan(v, u);
      } else
        low[u] = low[v] = true;
    }
  }
}
int lca(int u, int v) {
  if (dep[u] > dep[v]) swap(u, v);
  int d = dep[v] - dep[u];
  for (register int i = 0; i < 20; i++)
    if (d >> i & 1) v = par[v][i];
  if (u == v) return v;
  for (register int i = 19; i >= 0; i--) {
    if (par[u][i] ^ par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  }
  return par[u][0];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0, x, y, cost; i < m; i++) {
    scanf("%d%d%d", &x, &y, &cost);
    g[x].push_back(make_pair(y, cost));
    g[y].push_back(make_pair(x, cost));
  }
  par[1][0] = 1;
  tarjan(1, 1);
  for (register int j = 1; j < 20; j++)
    for (register int i = 1; i <= n; i++) par[i][j] = par[par[i][j - 1]][j - 1];
  for (register int i = 1; i <= n; i++)
    if (low[i]) qf.push_back(i);
  for (register int num = 0; num < qf.size(); num++) {
    for (register int i = 1; i <= n; i++) dist[i][num] = lnf;
    priority_queue<pair<long long, int> > pq;
    dist[qf[num]][num] = 0;
    pq.push(make_pair(0, qf[num]));
    while (!pq.empty()) {
      pair<long long, int> cur = pq.top();
      pq.pop();
      if (dist[cur.second][num] ^ -cur.first) continue;
      int u = cur.second;
      for (register int i = 0; i < g[u].size(); i++) {
        int v = g[u][i].first;
        if (dist[v][num] > dist[u][num] + g[u][i].second) {
          dist[v][num] = dist[u][num] + g[u][i].second;
          pq.push(make_pair(-dist[v][num], v));
        }
      }
    }
  }
  scanf("%d", &q);
  for (int u, v, lc; q--;) {
    scanf("%d%d", &u, &v);
    lc = lca(u, v);
    long long ans = sum[u] + sum[v] - 2 * sum[lc];
    for (int i = 0; i < qf.size(); i++) ans = min(ans, dist[u][i] + dist[v][i]);
    printf("%lld\n", ans);
  }
  return 0;
}

```