#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MAXN = (int)4e5 + 5;
long long n, ans[MAXN], ext[MAXN];
vector<pair<int, pair<int, int>>> sorg[MAXN];
vector<pair<int, int>> adj[MAXN];
int a[MAXN];
const int L = 20;
int geri[MAXN];
int par[MAXN][L], dep[MAXN];
int tin[MAXN], tout[MAXN];
int Gt = 1;
void dfs0(int v, int pr) {
  tin[v] = Gt++;
  if (pr == -1) {
    dep[v] = 0;
    par[v][0] = v;
  } else {
    dep[v] = dep[pr] + 1;
    par[v][0] = pr;
  }
  for (int j = 1; j < L; j++) par[v][j] = par[par[v][j - 1]][j - 1];
  ext[v] = a[v];
  for (pair<int, int> j : adj[v]) {
    if (j.first == pr) continue;
    geri[j.first] = j.second;
    dfs0(j.first, v);
    ext[v] += max(0LL, -2LL * j.second + ext[j.first]);
  }
  tout[v] = Gt - 1;
}
int gtp(int v, int dd) {
  for (int i = L - 1; i >= 0; i--) {
    if ((1 << i) <= dd) {
      v = par[v][i];
      dd -= (1 << i);
    }
  }
  return v;
}
int find_lca(int a, int b) {
  if (dep[a] > dep[b]) swap(a, b);
  for (int i = L - 1; i >= 0; i--) {
    if (dep[a] + (1 << i) <= dep[b]) b = par[b][i];
  }
  if (a == b) return a;
  for (int i = L - 1; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
    }
  }
  return par[a][0];
}
long long t[MAXN];
void upd(int l, int r, long long val) {
  r++;
  for (; l < MAXN; l += (l & -l)) t[l] += val;
  for (; r < MAXN; r += (r & -r)) t[r] -= val;
}
long long gt(int v) {
  long long res = 0;
  for (; v; v -= (v & -v)) res += t[v];
  return res;
}
void dfs1(int v, int pr, long long sum) {
  upd(tin[v], tin[v], ext[v]);
  long long nS = sum + a[v];
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    nS += max(0LL, -2 * u.second + ext[u.first]);
  }
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    nS -= max(0LL, -2 * u.second + ext[u.first]);
    dfs1(u.first, v, max(0LL, nS - 2 * u.second));
    nS += max(0LL, -2 * u.second + ext[u.first]);
  }
  unordered_map<int, long long> mp;
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    long long dg = max(0LL, -2 * u.second + ext[u.first]);
    mp[u.first] = dg;
  }
  for (auto u : sorg[v]) {
    int a = u.second.first, b = u.second.second;
    if (b == v) swap(a, b);
    if (a == v && b == v) {
      ans[u.first] = sum + ext[v];
    } else if (a == v) {
      int r1 = gtp(b, dep[b] - dep[v] - 1);
      ans[u.first] = sum + ext[v] + gt(tin[b]) - mp[r1] - geri[r1];
    } else {
      int r1 = gtp(b, dep[b] - dep[v] - 1), r2 = gtp(a, dep[a] - dep[v] - 1);
      ans[u.first] = sum + ext[v] + gt(tin[b]) + gt(tin[a]) - mp[r1] - mp[r2] -
                     geri[r1] - geri[r2];
    }
  }
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    upd(tin[u.first], tout[u.first], ext[v] - mp[u.first] - u.second);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  dfs0(1, -1);
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    sorg[find_lca(u, v)].push_back({i, {u, v}});
  }
  dfs1(1, -1, 0);
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}

### Prompt

Please create a solution in cpp to the following problem:

You are given a weighted tree (undirected connected graph with no cycles, loops or multiple edges) with n vertices. The edge \\{u_j, v_j\} has weight w_j. Also each vertex i has its own value a_i assigned to it.

Let's call a path starting in vertex u and ending in vertex v, where each edge can appear no more than twice (regardless of direction), a 2-path. Vertices can appear in the 2-path multiple times (even start and end vertices).

For some 2-path p profit Pr(p) = ∑_{v ∈ distinct vertices in  p}{a_v} - ∑_{e ∈ distinct edges in  p}{k_e ⋅ w_e}, where k_e is the number of times edge e appears in p. That is, vertices are counted once, but edges are counted the number of times they appear in p.

You are about to answer m queries. Each query is a pair of vertices (qu, qv). For each query find 2-path p from qu to qv with maximal profit Pr(p).

Input

The first line contains two integers n and q (2 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ q ≤ 4 ⋅ 10^5) — the number of vertices in the tree and the number of queries.

The second line contains n space-separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the values of the vertices.

Next n - 1 lines contain descriptions of edges: each line contains three space separated integers u_i, v_i and w_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i, 1 ≤ w_i ≤ 10^9) — there is edge \\{u_i, v_i\} with weight w_i in the tree.

Next q lines contain queries (one per line). Each query contains two integers qu_i and qv_i (1 ≤ qu_i, qv_i ≤ n) — endpoints of the 2-path you need to find.

Output

For each query print one integer per line — maximal profit Pr(p) of the some 2-path p with the corresponding endpoints.

Example

Input

7 6
6 5 5 3 2 1 2
1 2 2
2 3 2
2 4 1
4 5 1
6 4 2
7 3 25
1 1
4 4
5 6
6 4
3 4
3 7


Output

9
9
9
8
12
-14

Note

Explanation of queries: 

  1. (1, 1) — one of the optimal 2-paths is the following: 1 → 2 → 4 → 5 → 4 → 2 → 3 → 2 → 1. Pr(p) = (a_1 + a_2 + a_3 + a_4 + a_5) - (2 ⋅ w(1,2) + 2 ⋅ w(2,3) + 2 ⋅ w(2,4) + 2 ⋅ w(4,5)) = 21 - 2 ⋅ 12 = 9. 
  2. (4, 4): 4 → 2 → 1 → 2 → 3 → 2 → 4. Pr(p) = (a_1 + a_2 + a_3 + a_4) - 2 ⋅ (w(1,2) + w(2,3) + w(2,4)) = 19 - 2 ⋅ 10 = 9. 
  3. (5, 6): 5 → 4 → 2 → 3 → 2 → 1 → 2 → 4 → 6. 
  4. (6, 4): 6 → 4 → 2 → 1 → 2 → 3 → 2 → 4. 
  5. (3, 4): 3 → 2 → 1 → 2 → 4. 
  6. (3, 7): 3 → 2 → 1 → 2 → 4 → 5 → 4 → 2 → 3 → 7. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1e18;
const int mod = (int)1e9 + 7;
const int MAXN = (int)4e5 + 5;
long long n, ans[MAXN], ext[MAXN];
vector<pair<int, pair<int, int>>> sorg[MAXN];
vector<pair<int, int>> adj[MAXN];
int a[MAXN];
const int L = 20;
int geri[MAXN];
int par[MAXN][L], dep[MAXN];
int tin[MAXN], tout[MAXN];
int Gt = 1;
void dfs0(int v, int pr) {
  tin[v] = Gt++;
  if (pr == -1) {
    dep[v] = 0;
    par[v][0] = v;
  } else {
    dep[v] = dep[pr] + 1;
    par[v][0] = pr;
  }
  for (int j = 1; j < L; j++) par[v][j] = par[par[v][j - 1]][j - 1];
  ext[v] = a[v];
  for (pair<int, int> j : adj[v]) {
    if (j.first == pr) continue;
    geri[j.first] = j.second;
    dfs0(j.first, v);
    ext[v] += max(0LL, -2LL * j.second + ext[j.first]);
  }
  tout[v] = Gt - 1;
}
int gtp(int v, int dd) {
  for (int i = L - 1; i >= 0; i--) {
    if ((1 << i) <= dd) {
      v = par[v][i];
      dd -= (1 << i);
    }
  }
  return v;
}
int find_lca(int a, int b) {
  if (dep[a] > dep[b]) swap(a, b);
  for (int i = L - 1; i >= 0; i--) {
    if (dep[a] + (1 << i) <= dep[b]) b = par[b][i];
  }
  if (a == b) return a;
  for (int i = L - 1; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
    }
  }
  return par[a][0];
}
long long t[MAXN];
void upd(int l, int r, long long val) {
  r++;
  for (; l < MAXN; l += (l & -l)) t[l] += val;
  for (; r < MAXN; r += (r & -r)) t[r] -= val;
}
long long gt(int v) {
  long long res = 0;
  for (; v; v -= (v & -v)) res += t[v];
  return res;
}
void dfs1(int v, int pr, long long sum) {
  upd(tin[v], tin[v], ext[v]);
  long long nS = sum + a[v];
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    nS += max(0LL, -2 * u.second + ext[u.first]);
  }
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    nS -= max(0LL, -2 * u.second + ext[u.first]);
    dfs1(u.first, v, max(0LL, nS - 2 * u.second));
    nS += max(0LL, -2 * u.second + ext[u.first]);
  }
  unordered_map<int, long long> mp;
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    long long dg = max(0LL, -2 * u.second + ext[u.first]);
    mp[u.first] = dg;
  }
  for (auto u : sorg[v]) {
    int a = u.second.first, b = u.second.second;
    if (b == v) swap(a, b);
    if (a == v && b == v) {
      ans[u.first] = sum + ext[v];
    } else if (a == v) {
      int r1 = gtp(b, dep[b] - dep[v] - 1);
      ans[u.first] = sum + ext[v] + gt(tin[b]) - mp[r1] - geri[r1];
    } else {
      int r1 = gtp(b, dep[b] - dep[v] - 1), r2 = gtp(a, dep[a] - dep[v] - 1);
      ans[u.first] = sum + ext[v] + gt(tin[b]) + gt(tin[a]) - mp[r1] - mp[r2] -
                     geri[r1] - geri[r2];
    }
  }
  for (pair<int, int> u : adj[v]) {
    if (u.first == pr) continue;
    upd(tin[u.first], tout[u.first], ext[v] - mp[u.first] - u.second);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({v, w});
    adj[v].push_back({u, w});
  }
  dfs0(1, -1);
  for (int i = 0; i < q; i++) {
    int u, v;
    cin >> u >> v;
    sorg[find_lca(u, v)].push_back({i, {u, v}});
  }
  dfs1(1, -1, 0);
  for (int i = 0; i < q; i++) cout << ans[i] << '\n';
}

```