#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10;
long long n, m, q, dep[N], Dis[N], dis[45][N], head[N], cnt = -1;
long long fa[N][20];
struct edge {
  long long to, nxt, w;
} e[N << 1];
vector<long long> vec;
bool vis[N], st[N << 1], inq[N];
inline void add(long long u, long long v, long long w) {
  e[++cnt].to = v;
  e[cnt].w = w;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long f) {
  dep[u] = dep[f] + 1;
  fa[u][0] = f;
  vis[u] = true;
  for (long long i = 1; i <= 18; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (long long i = head[u]; ~i; i = e[i].nxt) {
    long long v = e[i].to;
    if (v == f || vis[v]) continue;
    Dis[v] = Dis[u] + e[i].w;
    st[i] = st[i ^ 1] = true;
    dfs(v, u);
  }
}
inline long long lca(long long x, long long y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (long long i = 18; i >= 0; i--)
    if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];
  if (x == y) return x;
  for (long long i = 18; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
inline void dij(long long s, long long t) {
  memset(vis, 0, sizeof(vis));
  memset(dis[t], 0x3f, sizeof(dis[t]));
  priority_queue<pair<long long, long long> > Q;
  dis[t][s] = 0;
  Q.push({0, s});
  while (!Q.empty()) {
    long long u = Q.top().second;
    Q.pop();
    if (vis[u]) continue;
    vis[u] = true;
    for (long long i = head[u]; ~i; i = e[i].nxt) {
      long long v = e[i].to;
      if (dis[t][v] > dis[t][u] + e[i].w) {
        dis[t][v] = dis[t][u] + e[i].w;
        Q.push({-dis[t][v], v});
      }
    }
  }
}
signed main() {
  memset(head, -1, sizeof(head));
  scanf("%lld%lld", &n, &m);
  for (long long i = 1, x, y, z; i <= m; i++) {
    scanf("%lld%lld%lld", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dfs(1, 0);
  for (long long i = 0; i <= cnt; i += 2)
    if (!st[i]) {
      if (!inq[e[i].to]) vec.push_back(e[i].to), inq[e[i].to] = true;
      if (!inq[e[i ^ 1].to])
        vec.push_back(e[i ^ 1].to), inq[e[i ^ 1].to] = true;
    }
  long long tot = 0;
  for (auto i : vec) dij(i, ++tot);
  scanf("%lld", &q);
  while (q--) {
    long long x, y, ans = 0;
    scanf("%lld%lld", &x, &y);
    ans = Dis[x] + Dis[y] - 2 * Dis[lca(x, y)];
    for (long long i = 1; i <= tot; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    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;
const long long N = 1e5 + 10;
long long n, m, q, dep[N], Dis[N], dis[45][N], head[N], cnt = -1;
long long fa[N][20];
struct edge {
  long long to, nxt, w;
} e[N << 1];
vector<long long> vec;
bool vis[N], st[N << 1], inq[N];
inline void add(long long u, long long v, long long w) {
  e[++cnt].to = v;
  e[cnt].w = w;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long f) {
  dep[u] = dep[f] + 1;
  fa[u][0] = f;
  vis[u] = true;
  for (long long i = 1; i <= 18; i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (long long i = head[u]; ~i; i = e[i].nxt) {
    long long v = e[i].to;
    if (v == f || vis[v]) continue;
    Dis[v] = Dis[u] + e[i].w;
    st[i] = st[i ^ 1] = true;
    dfs(v, u);
  }
}
inline long long lca(long long x, long long y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (long long i = 18; i >= 0; i--)
    if (dep[fa[x][i]] >= dep[y]) x = fa[x][i];
  if (x == y) return x;
  for (long long i = 18; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
inline void dij(long long s, long long t) {
  memset(vis, 0, sizeof(vis));
  memset(dis[t], 0x3f, sizeof(dis[t]));
  priority_queue<pair<long long, long long> > Q;
  dis[t][s] = 0;
  Q.push({0, s});
  while (!Q.empty()) {
    long long u = Q.top().second;
    Q.pop();
    if (vis[u]) continue;
    vis[u] = true;
    for (long long i = head[u]; ~i; i = e[i].nxt) {
      long long v = e[i].to;
      if (dis[t][v] > dis[t][u] + e[i].w) {
        dis[t][v] = dis[t][u] + e[i].w;
        Q.push({-dis[t][v], v});
      }
    }
  }
}
signed main() {
  memset(head, -1, sizeof(head));
  scanf("%lld%lld", &n, &m);
  for (long long i = 1, x, y, z; i <= m; i++) {
    scanf("%lld%lld%lld", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dfs(1, 0);
  for (long long i = 0; i <= cnt; i += 2)
    if (!st[i]) {
      if (!inq[e[i].to]) vec.push_back(e[i].to), inq[e[i].to] = true;
      if (!inq[e[i ^ 1].to])
        vec.push_back(e[i ^ 1].to), inq[e[i ^ 1].to] = true;
    }
  long long tot = 0;
  for (auto i : vec) dij(i, ++tot);
  scanf("%lld", &q);
  while (q--) {
    long long x, y, ans = 0;
    scanf("%lld%lld", &x, &y);
    ans = Dis[x] + Dis[y] - 2 * Dis[lca(x, y)];
    for (long long i = 1; i <= tot; i++) ans = min(ans, dis[i][x] + dis[i][y]);
    printf("%lld\n", ans);
  }
  return 0;
}

```