#include <bits/stdc++.h>
using namespace std;
int a[300009], p[300009][19], lvl[300009], n;
long long int upsum[300009][19], dpdown[300009], dpup[300009], paredge[300009];
vector<pair<int, int> > l[300009];
void dfs(int ver, int par) {
  dpdown[ver] = a[ver];
  for (auto it : l[ver]) {
    if (it.first == par) {
      continue;
    }
    lvl[it.first] = lvl[ver] + 1;
    p[it.first][0] = ver;
    dfs(it.first, ver);
    dpdown[ver] += max(0ll, dpdown[it.first] - 2ll * it.second);
  }
  return;
}
void dfs2(int ver, int par, int curedge) {
  int i, j, t1, t2, t3, t4;
  long long int buf1, buf2, buf3, buf4;
  paredge[ver] = curedge;
  if (par != -1) {
    buf1 = max(0ll, dpdown[ver] - 2ll * curedge);
    upsum[ver][0] = dpdown[par] - buf1 - curedge;
    buf1 = dpup[par] + (dpdown[par] - buf1) - 2ll * curedge;
    buf1 = max(0ll, buf1);
    dpup[ver] = buf1;
  }
  for (auto it : l[ver]) {
    if (it.first == par) {
      continue;
    }
    dfs2(it.first, ver, it.second);
  }
  return;
}
int lca(int u, int v) {
  int lg, i;
  if (lvl[u] < lvl[v]) {
    swap(u, v);
  }
  for (lg = 0; (1 << lg) <= lvl[u]; lg++) {
  }
  lg--;
  for (i = lg; i >= 0; i--) {
    if (lvl[u] - (1 << i) >= lvl[v]) {
      u = p[u][i];
    }
  }
  if (u == v) {
    return u;
  }
  for (i = lg; i >= 0; i--) {
    if (p[u][i] != -1 && p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  }
  return p[u][0];
}
void preprocess(int root) {
  int i, j, t1, t2;
  memset(p, -1, sizeof(p));
  lvl[root] = 0;
  dfs(root, -1);
  dfs2(root, -1, -1);
  for (i = 1; i < 19; i++) {
    for (j = 1; j <= n; j++) {
      if (p[j][i - 1] != -1) {
        p[j][i] = p[p[j][i - 1]][i - 1];
        upsum[j][i] = upsum[j][i - 1] + upsum[p[j][i - 1]][i - 1];
      }
    }
  }
  return;
}
long long int solve(int u, int v) {
  int i, j, t1, t2, t3, t4, heightdiff, anc, cur;
  long long int buf1, buf2, buf3, buf4, tempans = 0;
  if (u == v) {
    return dpdown[u] + dpup[u];
  }
  if (lvl[u] > lvl[v]) {
    swap(u, v);
  }
  anc = lca(u, v);
  if (anc == u) {
    tempans = dpdown[v] + dpup[anc];
    cur = v;
    heightdiff = lvl[v] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    tempans += (dpdown[anc] - buf1);
    tempans -= paredge[cur];
  } else {
    tempans = dpdown[u] + dpdown[v] + dpup[anc];
    buf2 = dpdown[anc];
    cur = v;
    heightdiff = lvl[v] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    buf2 -= buf1;
    tempans -= paredge[cur];
    cur = u;
    heightdiff = lvl[u] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    buf2 -= buf1;
    tempans -= paredge[cur];
    tempans += buf2;
  }
  return tempans;
}
int main() {
  int i, j, t1, t2, t3, t4, q;
  long long int ans;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &t1, &t2, &t3);
    l[t1].push_back(make_pair(t2, t3));
    l[t2].push_back(make_pair(t1, t3));
  }
  preprocess(1);
  for (i = 0; i < q; i++) {
    scanf("%d %d", &t1, &t2);
    ans = solve(t1, t2);
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Generate a CPP solution 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;
int a[300009], p[300009][19], lvl[300009], n;
long long int upsum[300009][19], dpdown[300009], dpup[300009], paredge[300009];
vector<pair<int, int> > l[300009];
void dfs(int ver, int par) {
  dpdown[ver] = a[ver];
  for (auto it : l[ver]) {
    if (it.first == par) {
      continue;
    }
    lvl[it.first] = lvl[ver] + 1;
    p[it.first][0] = ver;
    dfs(it.first, ver);
    dpdown[ver] += max(0ll, dpdown[it.first] - 2ll * it.second);
  }
  return;
}
void dfs2(int ver, int par, int curedge) {
  int i, j, t1, t2, t3, t4;
  long long int buf1, buf2, buf3, buf4;
  paredge[ver] = curedge;
  if (par != -1) {
    buf1 = max(0ll, dpdown[ver] - 2ll * curedge);
    upsum[ver][0] = dpdown[par] - buf1 - curedge;
    buf1 = dpup[par] + (dpdown[par] - buf1) - 2ll * curedge;
    buf1 = max(0ll, buf1);
    dpup[ver] = buf1;
  }
  for (auto it : l[ver]) {
    if (it.first == par) {
      continue;
    }
    dfs2(it.first, ver, it.second);
  }
  return;
}
int lca(int u, int v) {
  int lg, i;
  if (lvl[u] < lvl[v]) {
    swap(u, v);
  }
  for (lg = 0; (1 << lg) <= lvl[u]; lg++) {
  }
  lg--;
  for (i = lg; i >= 0; i--) {
    if (lvl[u] - (1 << i) >= lvl[v]) {
      u = p[u][i];
    }
  }
  if (u == v) {
    return u;
  }
  for (i = lg; i >= 0; i--) {
    if (p[u][i] != -1 && p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  }
  return p[u][0];
}
void preprocess(int root) {
  int i, j, t1, t2;
  memset(p, -1, sizeof(p));
  lvl[root] = 0;
  dfs(root, -1);
  dfs2(root, -1, -1);
  for (i = 1; i < 19; i++) {
    for (j = 1; j <= n; j++) {
      if (p[j][i - 1] != -1) {
        p[j][i] = p[p[j][i - 1]][i - 1];
        upsum[j][i] = upsum[j][i - 1] + upsum[p[j][i - 1]][i - 1];
      }
    }
  }
  return;
}
long long int solve(int u, int v) {
  int i, j, t1, t2, t3, t4, heightdiff, anc, cur;
  long long int buf1, buf2, buf3, buf4, tempans = 0;
  if (u == v) {
    return dpdown[u] + dpup[u];
  }
  if (lvl[u] > lvl[v]) {
    swap(u, v);
  }
  anc = lca(u, v);
  if (anc == u) {
    tempans = dpdown[v] + dpup[anc];
    cur = v;
    heightdiff = lvl[v] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    tempans += (dpdown[anc] - buf1);
    tempans -= paredge[cur];
  } else {
    tempans = dpdown[u] + dpdown[v] + dpup[anc];
    buf2 = dpdown[anc];
    cur = v;
    heightdiff = lvl[v] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    buf2 -= buf1;
    tempans -= paredge[cur];
    cur = u;
    heightdiff = lvl[u] - lvl[anc] - 1;
    for (i = 18; i >= 0; i--) {
      if (heightdiff & (1 << i)) {
        tempans += upsum[cur][i];
        cur = p[cur][i];
      }
    }
    buf1 = max(0ll, dpdown[cur] - 2 * paredge[cur]);
    buf2 -= buf1;
    tempans -= paredge[cur];
    tempans += buf2;
  }
  return tempans;
}
int main() {
  int i, j, t1, t2, t3, t4, q;
  long long int ans;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &t1, &t2, &t3);
    l[t1].push_back(make_pair(t2, t3));
    l[t2].push_back(make_pair(t1, t3));
  }
  preprocess(1);
  for (i = 0; i < q; i++) {
    scanf("%d %d", &t1, &t2);
    ans = solve(t1, t2);
    printf("%lld\n", ans);
  }
  return 0;
}

```