#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXINT = 2147483647;
const int N = 300 * 1000 + 9;
const int LOGN = 19;
const int M = 21;
const long long INF64 = 1e18;
int n, m, q;
vector<pair<int, int> > g[N];
int p[LOGN][N];
int tin[N], tout[N], T;
long long h[N];
set<pair<int, int> > badEdges;
long long d[M + M][N];
bool u[N];
void dfs(int v, int pr) {
  tin[v] = T++;
  p[0][v] = pr;
  u[v] = true;
  for (int i = 1; i < LOGN; ++i) p[i][v] = p[i - 1][p[i - 1][v]];
  vector<pair<int, int> >::iterator e;
  for (e = g[v].begin(); e != g[v].end(); e++) {
    int to = e->first, len = e->second;
    if (!u[to]) {
      h[to] = h[v] + len;
      dfs(to, v);
      if (v < to)
        badEdges.erase(make_pair(v, to));
      else
        badEdges.erase(make_pair(to, v));
    }
  }
  tout[v] = T;
}
bool isAncestor(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int getLCA(int a, int b) {
  if (isAncestor(a, b)) return a;
  if (isAncestor(b, a)) return b;
  for (int i = LOGN - 1; i >= 0; --i)
    if (!isAncestor(p[i][a], b)) a = p[i][a];
  return p[0][a];
}
void dij(int st, long long d[N]) {
  set<pair<long long, int> > q;
  for (int i = 0; i < n; ++i) d[i] = INF64;
  d[st] = 0;
  q.insert(make_pair(d[st], st));
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    vector<pair<int, int> >::iterator e;
    for (e = g[v].begin(); e != g[v].end(); e++) {
      int to = e->first, len = e->second;
      if (d[to] > d[v] + len) {
        q.erase(make_pair(d[to], to));
        d[to] = d[v] + len;
        q.insert(make_pair(d[to], to));
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    --u, --v;
    g[u].push_back(make_pair(v, w));
    g[v].push_back(make_pair(u, w));
  }
  for (int v = 0; v < n; ++v)
    for (vector<pair<int, int> >::iterator e = g[v].begin(); e != g[v].end();
         e++)
      if (v < e->first) badEdges.insert(make_pair(v, e->first));
  dfs(0, 0);
  int cpos = 0;
  for (set<pair<int, int> >::iterator e = badEdges.begin(); e != badEdges.end();
       e++)
    dij(e->first, d[cpos++]);
  scanf("%d", &q);
  for (int tc = 0; tc < q; ++tc) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    int lca = getLCA(u, v);
    long long ans = h[u] + h[v] - 2 * h[lca];
    for (int i = 0; i < badEdges.size(); ++i) ans = min(ans, d[i][u] + d[i][v]);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 INF = 0x3f3f3f3f;
const int MAXINT = 2147483647;
const int N = 300 * 1000 + 9;
const int LOGN = 19;
const int M = 21;
const long long INF64 = 1e18;
int n, m, q;
vector<pair<int, int> > g[N];
int p[LOGN][N];
int tin[N], tout[N], T;
long long h[N];
set<pair<int, int> > badEdges;
long long d[M + M][N];
bool u[N];
void dfs(int v, int pr) {
  tin[v] = T++;
  p[0][v] = pr;
  u[v] = true;
  for (int i = 1; i < LOGN; ++i) p[i][v] = p[i - 1][p[i - 1][v]];
  vector<pair<int, int> >::iterator e;
  for (e = g[v].begin(); e != g[v].end(); e++) {
    int to = e->first, len = e->second;
    if (!u[to]) {
      h[to] = h[v] + len;
      dfs(to, v);
      if (v < to)
        badEdges.erase(make_pair(v, to));
      else
        badEdges.erase(make_pair(to, v));
    }
  }
  tout[v] = T;
}
bool isAncestor(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int getLCA(int a, int b) {
  if (isAncestor(a, b)) return a;
  if (isAncestor(b, a)) return b;
  for (int i = LOGN - 1; i >= 0; --i)
    if (!isAncestor(p[i][a], b)) a = p[i][a];
  return p[0][a];
}
void dij(int st, long long d[N]) {
  set<pair<long long, int> > q;
  for (int i = 0; i < n; ++i) d[i] = INF64;
  d[st] = 0;
  q.insert(make_pair(d[st], st));
  while (!q.empty()) {
    int v = q.begin()->second;
    q.erase(q.begin());
    vector<pair<int, int> >::iterator e;
    for (e = g[v].begin(); e != g[v].end(); e++) {
      int to = e->first, len = e->second;
      if (d[to] > d[v] + len) {
        q.erase(make_pair(d[to], to));
        d[to] = d[v] + len;
        q.insert(make_pair(d[to], to));
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    --u, --v;
    g[u].push_back(make_pair(v, w));
    g[v].push_back(make_pair(u, w));
  }
  for (int v = 0; v < n; ++v)
    for (vector<pair<int, int> >::iterator e = g[v].begin(); e != g[v].end();
         e++)
      if (v < e->first) badEdges.insert(make_pair(v, e->first));
  dfs(0, 0);
  int cpos = 0;
  for (set<pair<int, int> >::iterator e = badEdges.begin(); e != badEdges.end();
       e++)
    dij(e->first, d[cpos++]);
  scanf("%d", &q);
  for (int tc = 0; tc < q; ++tc) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    int lca = getLCA(u, v);
    long long ans = h[u] + h[v] - 2 * h[lca];
    for (int i = 0; i < badEdges.size(); ++i) ans = min(ans, d[i][u] + d[i][v]);
    printf("%I64d\n", ans);
  }
  return 0;
}

```