#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct edg {
  int to, nxt, fro;
  long long w;
  bool is;
} e[N << 1];
struct Node {
  int v;
  long long w;
  Node(){};
  Node(int _v, long long _w) : v(_v), w(_w){};
};
bool operator<(Node x, Node y) { return x.w > y.w; }
int n, m, cnt = 1, head[N], f[N], id[N], num, ro, dfn[N], que;
long long dis[50][N], tdis[N], dep[N], fa[N][25];
bool vis[N];
priority_queue<Node> q;
void add(int u, int v, long long w) {
  cnt++;
  e[cnt].fro = u;
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
  e[cnt].w = w;
}
void diji(int s) {
  memset(dis[id[s]], 0x3f, sizeof(dis[id[s]]));
  memset(vis, 0, sizeof(vis));
  dis[id[s]][s] = 0;
  q.push(Node(s, 0));
  while (q.size()) {
    int u = q.top().v;
    q.pop();
    if (vis[u]) continue;
    vis[u] = 1;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (dis[id[s]][u] + e[i].w < dis[id[s]][v]) {
        dis[id[s]][v] = dis[id[s]][u] + e[i].w;
        q.push(Node(v, dis[id[s]][v]));
      }
    }
  }
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (vis[v]) continue;
    dep[v] = dep[u] + 1;
    fa[v][0] = u;
    tdis[v] = e[i].w + tdis[u];
    e[i].is = e[i ^ 1].is = 1;
    dfs(v);
  }
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int d = dep[u] - dep[v];
  for (int i = 20; i >= 0; i--)
    if (d & (1 << i)) u = fa[u][i];
  if (u == v) return u;
  for (int i = 20; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) {
      u = fa[u][i];
      v = fa[v][i];
    }
  return fa[u][0];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    long long w;
    scanf("%d%d%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  dfs(1);
  for (int j = 1; j <= 20; j++)
    for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  for (int i = 2; i <= cnt; i++) {
    if (e[i].is) continue;
    if (!id[e[i].fro]) {
      id[e[i].fro] = ++num;
      dfn[num] = e[i].fro;
      diji(e[i].fro);
    }
    if (!id[e[i].to]) {
      id[e[i].to] = ++num;
      dfn[num] = e[i].to;
      diji(e[i].to);
    }
  }
  scanf("%d", &que);
  while (que--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int lca = getlca(x, y);
    long long ans = tdis[x] + tdis[y] - 2 * tdis[lca];
    for (int i = 1; i <= num; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    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 N = 2e5 + 5;
struct edg {
  int to, nxt, fro;
  long long w;
  bool is;
} e[N << 1];
struct Node {
  int v;
  long long w;
  Node(){};
  Node(int _v, long long _w) : v(_v), w(_w){};
};
bool operator<(Node x, Node y) { return x.w > y.w; }
int n, m, cnt = 1, head[N], f[N], id[N], num, ro, dfn[N], que;
long long dis[50][N], tdis[N], dep[N], fa[N][25];
bool vis[N];
priority_queue<Node> q;
void add(int u, int v, long long w) {
  cnt++;
  e[cnt].fro = u;
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
  e[cnt].w = w;
}
void diji(int s) {
  memset(dis[id[s]], 0x3f, sizeof(dis[id[s]]));
  memset(vis, 0, sizeof(vis));
  dis[id[s]][s] = 0;
  q.push(Node(s, 0));
  while (q.size()) {
    int u = q.top().v;
    q.pop();
    if (vis[u]) continue;
    vis[u] = 1;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (dis[id[s]][u] + e[i].w < dis[id[s]][v]) {
        dis[id[s]][v] = dis[id[s]][u] + e[i].w;
        q.push(Node(v, dis[id[s]][v]));
      }
    }
  }
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (vis[v]) continue;
    dep[v] = dep[u] + 1;
    fa[v][0] = u;
    tdis[v] = e[i].w + tdis[u];
    e[i].is = e[i ^ 1].is = 1;
    dfs(v);
  }
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int d = dep[u] - dep[v];
  for (int i = 20; i >= 0; i--)
    if (d & (1 << i)) u = fa[u][i];
  if (u == v) return u;
  for (int i = 20; i >= 0; i--)
    if (fa[u][i] != fa[v][i]) {
      u = fa[u][i];
      v = fa[v][i];
    }
  return fa[u][0];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    long long w;
    scanf("%d%d%lld", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  dfs(1);
  for (int j = 1; j <= 20; j++)
    for (int i = 1; i <= n; i++) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  for (int i = 2; i <= cnt; i++) {
    if (e[i].is) continue;
    if (!id[e[i].fro]) {
      id[e[i].fro] = ++num;
      dfn[num] = e[i].fro;
      diji(e[i].fro);
    }
    if (!id[e[i].to]) {
      id[e[i].to] = ++num;
      dfn[num] = e[i].to;
      diji(e[i].to);
    }
  }
  scanf("%d", &que);
  while (que--) {
    int x, y;
    scanf("%d%d", &x, &y);
    int lca = getlca(x, y);
    long long ans = tdis[x] + tdis[y] - 2 * tdis[lca];
    for (int i = 1; i <= num; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    printf("%lld\n", ans);
  }
  return 0;
}

```