#include <bits/stdc++.h>
using namespace std;
const int mod = 1000 * 1000 * 1000 + 7;
const int N = 3000 * 100 + 10, M = 50 + 5, K = 20 + 2;
vector<pair<int, int>> adj[N];
long long dis[M][N], d[N], h[N];
int n, m, id;
int par[N][K];
bool mark[N];
void dij(int x) {
  set<pair<long long, int>> s;
  for (int i = 1; i <= n; i++) dis[id][i] = 1e18;
  dis[id][x] = 0;
  s.insert({0, x});
  while ((int)s.size()) {
    int x = s.begin()->second;
    s.erase(s.begin());
    for (auto u : adj[x]) {
      if (dis[id][u.first] > dis[id][x] + u.second) {
        s.erase({dis[id][u.first], u.first});
        dis[id][u.first] = dis[id][x] + u.second;
        s.insert({dis[id][u.first], u.first});
      }
    }
  }
  id++;
}
void dfs(int u, int p) {
  mark[u] = 1;
  par[u][0] = p;
  for (int i = 1; i < K; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  for (auto v : adj[u]) {
    if (v.first == p) continue;
    if (mark[v.first]) {
      dij(u);
      continue;
    }
    d[v.first] = d[u] + v.second;
    h[v.first] = h[u] + 1;
    dfs(v.first, u);
  }
}
int ipar(int x, int y) {
  for (int i = 0; i < K; i++)
    if (y & (1 << i)) x = par[x][i];
  return x;
}
int lca(int x, int y) {
  if (h[x] > h[y]) swap(x, y);
  y = ipar(y, h[y] - h[x]);
  if (x == y) return x;
  for (int i = K - 1; i > -1; i--)
    if (par[x][i] != par[y][i]) x = par[x][i], y = par[y][i];
  return par[x][0];
}
long long dist(int x, int y) { return d[x] + d[y] - 2 * d[lca(x, y)]; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    adj[x].push_back({y, w});
    adj[y].push_back({x, w});
  }
  dfs(1, -1);
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    long long ans = dist(x, y);
    for (int i = 0; i < id; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    cout << ans << '\n';
  }
}

### Prompt

Create a solution in CPP for 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 mod = 1000 * 1000 * 1000 + 7;
const int N = 3000 * 100 + 10, M = 50 + 5, K = 20 + 2;
vector<pair<int, int>> adj[N];
long long dis[M][N], d[N], h[N];
int n, m, id;
int par[N][K];
bool mark[N];
void dij(int x) {
  set<pair<long long, int>> s;
  for (int i = 1; i <= n; i++) dis[id][i] = 1e18;
  dis[id][x] = 0;
  s.insert({0, x});
  while ((int)s.size()) {
    int x = s.begin()->second;
    s.erase(s.begin());
    for (auto u : adj[x]) {
      if (dis[id][u.first] > dis[id][x] + u.second) {
        s.erase({dis[id][u.first], u.first});
        dis[id][u.first] = dis[id][x] + u.second;
        s.insert({dis[id][u.first], u.first});
      }
    }
  }
  id++;
}
void dfs(int u, int p) {
  mark[u] = 1;
  par[u][0] = p;
  for (int i = 1; i < K; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  for (auto v : adj[u]) {
    if (v.first == p) continue;
    if (mark[v.first]) {
      dij(u);
      continue;
    }
    d[v.first] = d[u] + v.second;
    h[v.first] = h[u] + 1;
    dfs(v.first, u);
  }
}
int ipar(int x, int y) {
  for (int i = 0; i < K; i++)
    if (y & (1 << i)) x = par[x][i];
  return x;
}
int lca(int x, int y) {
  if (h[x] > h[y]) swap(x, y);
  y = ipar(y, h[y] - h[x]);
  if (x == y) return x;
  for (int i = K - 1; i > -1; i--)
    if (par[x][i] != par[y][i]) x = par[x][i], y = par[y][i];
  return par[x][0];
}
long long dist(int x, int y) { return d[x] + d[y] - 2 * d[lca(x, y)]; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    adj[x].push_back({y, w});
    adj[y].push_back({x, w});
  }
  dfs(1, -1);
  int q;
  cin >> q;
  while (q--) {
    int x, y;
    cin >> x >> y;
    long long ans = dist(x, y);
    for (int i = 0; i < id; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    cout << ans << '\n';
  }
}

```