#include <bits/stdc++.h>
using namespace std;
int n, m, q, f[200005], fa[200005][20], dep[200005];
long long dis[100][200005], d[200005];
vector<pair<int, int> > g[200005], gg[200005];
vector<int> vv, vvv;
struct edg {
  int u, v, w;
} e[200005];
int fnd(int x) { return f[x] == x ? x : f[x] = fnd(f[x]); }
void dfs(int x, int p) {
  fa[x][0] = p;
  dep[x] = dep[p] + 1;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i].first != p)
      d[g[x][i].first] = d[x] + g[x][i].second, dfs(g[x][i].first, x);
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int sub = dep[u] - dep[v];
  for (int i = 18; i >= 0; i--)
    if (sub & (1 << i)) u = fa[u][i];
  if (u == v) return u;
  for (int i = 18; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
void bfs(int id, int s) {
  priority_queue<pair<long long, int> > q;
  dis[id][s] = 0;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    int x = q.top().second;
    q.pop();
    for (int i = 0; i < gg[x].size(); i++)
      if (dis[id][gg[x][i].first] == -1 ||
          dis[id][gg[x][i].first] > dis[id][x] + gg[x][i].second) {
        dis[id][gg[x][i].first] = dis[id][x] + gg[x][i].second;
        q.push(make_pair(-dis[id][gg[x][i].first], gg[x][i].first));
      }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v, w; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    e[i] = (edg){u, v, w};
    gg[u].push_back(make_pair(v, w));
    gg[v].push_back(make_pair(u, w));
  }
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 1; i <= m; i++) {
    int f1 = fnd(e[i].u), f2 = fnd(e[i].v);
    if (f1 == f2) {
      vv.push_back(e[i].u), vv.push_back(e[i].v), vvv.push_back(e[i].w),
          vvv.push_back(e[i].w);
    } else {
      f[f1] = f2, g[e[i].u].push_back(make_pair(e[i].v, e[i].w)),
      g[e[i].v].push_back(make_pair(e[i].u, e[i].w));
    }
  }
  dfs(1, 0);
  for (int j = 1; j <= 18; j++)
    for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  memset(dis, -1, sizeof(dis));
  for (int i = 0; i < vv.size(); i++) bfs(i, vv[i]);
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    long long ans = d[x] + d[y] - 2 * d[lca(x, y)];
    for (int i = 0; i < vv.size(); i += 2)
      ans = min(ans, min(dis[i][x] + dis[i + 1][y], dis[i][y] + dis[i + 1][x]) +
                         vvv[i]);
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
int n, m, q, f[200005], fa[200005][20], dep[200005];
long long dis[100][200005], d[200005];
vector<pair<int, int> > g[200005], gg[200005];
vector<int> vv, vvv;
struct edg {
  int u, v, w;
} e[200005];
int fnd(int x) { return f[x] == x ? x : f[x] = fnd(f[x]); }
void dfs(int x, int p) {
  fa[x][0] = p;
  dep[x] = dep[p] + 1;
  for (int i = 0; i < g[x].size(); i++)
    if (g[x][i].first != p)
      d[g[x][i].first] = d[x] + g[x][i].second, dfs(g[x][i].first, x);
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int sub = dep[u] - dep[v];
  for (int i = 18; i >= 0; i--)
    if (sub & (1 << i)) u = fa[u][i];
  if (u == v) return u;
  for (int i = 18; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
void bfs(int id, int s) {
  priority_queue<pair<long long, int> > q;
  dis[id][s] = 0;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    int x = q.top().second;
    q.pop();
    for (int i = 0; i < gg[x].size(); i++)
      if (dis[id][gg[x][i].first] == -1 ||
          dis[id][gg[x][i].first] > dis[id][x] + gg[x][i].second) {
        dis[id][gg[x][i].first] = dis[id][x] + gg[x][i].second;
        q.push(make_pair(-dis[id][gg[x][i].first], gg[x][i].first));
      }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v, w; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    e[i] = (edg){u, v, w};
    gg[u].push_back(make_pair(v, w));
    gg[v].push_back(make_pair(u, w));
  }
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 1; i <= m; i++) {
    int f1 = fnd(e[i].u), f2 = fnd(e[i].v);
    if (f1 == f2) {
      vv.push_back(e[i].u), vv.push_back(e[i].v), vvv.push_back(e[i].w),
          vvv.push_back(e[i].w);
    } else {
      f[f1] = f2, g[e[i].u].push_back(make_pair(e[i].v, e[i].w)),
      g[e[i].v].push_back(make_pair(e[i].u, e[i].w));
    }
  }
  dfs(1, 0);
  for (int j = 1; j <= 18; j++)
    for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  memset(dis, -1, sizeof(dis));
  for (int i = 0; i < vv.size(); i++) bfs(i, vv[i]);
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    long long ans = d[x] + d[y] - 2 * d[lca(x, y)];
    for (int i = 0; i < vv.size(); i += 2)
      ans = min(ans, min(dis[i][x] + dis[i + 1][y], dis[i][y] + dis[i + 1][x]) +
                         vvv[i]);
    printf("%lld\n", ans);
  }
  return 0;
}

```