#include <bits/stdc++.h>
using namespace std;
const int N = 4 * 1e5 + 69;
long long b[N], res[N], in[N], out[N];
int p[N][20], h[N], n, q, val[N], edge[N];
vector<pair<int, int> > query[N], a[N];
void update(int first, long long val) {
  while (first <= n) {
    b[first] += val;
    first += first & -first;
  }
}
long long get(int first) {
  long long tmp = 0;
  while (first) {
    tmp += b[first];
    first -= first & -first;
  }
  return tmp;
}
int lca(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  int d = h[v] - h[u];
  for (int i = 19; i >= 0; i--)
    if ((d >> i) & 1) v = p[v][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; i--)
    if (p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  return p[u][0];
}
int nex(int u, int d) {
  if (d == -1) return -1;
  for (int i = 19; i >= 0; i--)
    if ((d >> i) & 1) u = p[u][i];
  return u;
}
void dfs1(int u) {
  in[u] = val[u];
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    h[v] = h[u] + 1;
    edge[v] = w;
    p[v][0] = u;
    for (int i = 1; i <= 19; i++) p[v][i] = p[p[v][i - 1]][i - 1];
    dfs1(v);
    in[u] += max(0LL, in[v] - 2 * w);
  }
}
void dfs2(int u) {
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    out[v] = val[v] + max(0LL, -2 * w + out[u] + in[u] -
                                   max(0LL, in[v] - 2 * w) - val[u]);
    dfs2(v);
  }
}
void calc(int u) {
  for (auto t : query[u]) {
    int v = t.first, id = t.second;
    res[id] += get(h[u]) - get(h[v] - 1) - edge[u] + in[u];
  }
  update(h[u], -edge[u] + in[u]);
  long long pre = 0;
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    update(h[u], -max(0LL, in[v] - 2 * w) - pre);
    pre = -max(0LL, in[v] - 2 * w);
    calc(v);
  }
  update(h[u], edge[u] - in[u] - pre);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    a[u].push_back(make_pair(v, w));
    a[v].push_back(make_pair(u, w));
  }
  h[1] = 1;
  dfs1(1);
  out[1] = val[1];
  dfs2(1);
  for (int i = 1; i <= q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    int p = lca(u, v);
    res[i] = out[p] + in[p] - val[p];
    int k = nex(u, h[u] - h[p] - 1);
    if (k != -1) {
      query[u].push_back(make_pair(k, i));
      res[i] -= max(0LL, in[k] - 2 * edge[k]);
    }
    k = nex(v, h[v] - h[p] - 1);
    if (k != -1) {
      query[v].push_back(make_pair(k, i));
      res[i] -= max(0LL, in[k] - 2 * edge[k]);
    }
  }
  calc(1);
  for (int i = 1; i <= q; i++) printf("%I64d\n", res[i]);
}

### Prompt

Please formulate 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;
const int N = 4 * 1e5 + 69;
long long b[N], res[N], in[N], out[N];
int p[N][20], h[N], n, q, val[N], edge[N];
vector<pair<int, int> > query[N], a[N];
void update(int first, long long val) {
  while (first <= n) {
    b[first] += val;
    first += first & -first;
  }
}
long long get(int first) {
  long long tmp = 0;
  while (first) {
    tmp += b[first];
    first -= first & -first;
  }
  return tmp;
}
int lca(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  int d = h[v] - h[u];
  for (int i = 19; i >= 0; i--)
    if ((d >> i) & 1) v = p[v][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; i--)
    if (p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  return p[u][0];
}
int nex(int u, int d) {
  if (d == -1) return -1;
  for (int i = 19; i >= 0; i--)
    if ((d >> i) & 1) u = p[u][i];
  return u;
}
void dfs1(int u) {
  in[u] = val[u];
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    h[v] = h[u] + 1;
    edge[v] = w;
    p[v][0] = u;
    for (int i = 1; i <= 19; i++) p[v][i] = p[p[v][i - 1]][i - 1];
    dfs1(v);
    in[u] += max(0LL, in[v] - 2 * w);
  }
}
void dfs2(int u) {
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    out[v] = val[v] + max(0LL, -2 * w + out[u] + in[u] -
                                   max(0LL, in[v] - 2 * w) - val[u]);
    dfs2(v);
  }
}
void calc(int u) {
  for (auto t : query[u]) {
    int v = t.first, id = t.second;
    res[id] += get(h[u]) - get(h[v] - 1) - edge[u] + in[u];
  }
  update(h[u], -edge[u] + in[u]);
  long long pre = 0;
  for (auto t : a[u]) {
    int v = t.first, w = t.second;
    if (v == p[u][0]) continue;
    update(h[u], -max(0LL, in[v] - 2 * w) - pre);
    pre = -max(0LL, in[v] - 2 * w);
    calc(v);
  }
  update(h[u], edge[u] - in[u] - pre);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) cin >> val[i];
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    a[u].push_back(make_pair(v, w));
    a[v].push_back(make_pair(u, w));
  }
  h[1] = 1;
  dfs1(1);
  out[1] = val[1];
  dfs2(1);
  for (int i = 1; i <= q; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    int p = lca(u, v);
    res[i] = out[p] + in[p] - val[p];
    int k = nex(u, h[u] - h[p] - 1);
    if (k != -1) {
      query[u].push_back(make_pair(k, i));
      res[i] -= max(0LL, in[k] - 2 * edge[k]);
    }
    k = nex(v, h[v] - h[p] - 1);
    if (k != -1) {
      query[v].push_back(make_pair(k, i));
      res[i] -= max(0LL, in[k] - 2 * edge[k]);
    }
  }
  calc(1);
  for (int i = 1; i <= q; i++) printf("%I64d\n", res[i]);
}

```