#include <bits/stdc++.h>
using std::max;
const int N = 3e5 + 5;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
struct line {
  int next, to, w;
} e[2 * N];
int n, q, hd[N], cnt, a[N], b[N], fa[N], dep[N], siz[N], hc[N], top[N];
long long s[N], f[N], g[N], h[N], c[N], ff[N];
void adline(int u, int v, int w) {
  e[cnt].to = v;
  e[cnt].w = w;
  e[cnt].next = hd[u];
  hd[u] = cnt++;
}
void dfs1(int u) {
  hc[u] = -1;
  s[u] = s[fa[u]] + a[u] - b[u];
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    fa[v] = u;
    b[v] = e[i].w;
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (!~hc[u] || siz[v] > siz[hc[u]]) hc[u] = v;
    f[u] += max(0ll, f[v] + a[v] - 2 * b[v]);
  }
  ++siz[u];
}
void dfs2(int u) {
  long long sum = (ff[u] = f[u]) + g[u] + a[u];
  h[u] = max(h[u], 0ll);
  if (~hc[u]) {
    int v = hc[u];
    top[v] = top[u];
    g[v] = max(0ll, sum - 2 * b[v] - max(0ll, h[v] = f[v] + a[v] - 2 * b[v]));
    dfs2(v);
  }
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u] || v == hc[u]) continue;
    g[v] = max(0ll, sum - 2 * b[v] - max(0ll, h[v] = f[v] + a[v] - 2 * b[v]));
    dfs2(top[v] = v);
  }
}
void dfs3(int u) {
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    f[v] += f[u];
    h[v] += h[u];
    dfs3(v);
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] > dep[top[v]])
      u = fa[top[u]];
    else
      v = fa[top[v]];
  }
  return dep[u] < dep[v] ? u : v;
}
int main() {
  read(n);
  read(q);
  for (int i = 0, i_end = n; i < i_end; ++i) read(a[i]);
  for (int i = 0, i_end = n; i < i_end; ++i) hd[i] = -1;
  for (int i = 0, i_end = n - 1; i < i_end; ++i) {
    int u, v, w;
    read(u);
    read(v);
    read(w);
    adline(--u, --v, w);
    adline(v, u, w);
  }
  dfs1(0);
  dfs2(0);
  dfs3(0);
  while (q--) {
    int u, v, m;
    long long ans;
    read(u);
    read(v);
    --u, --v;
    m = lca(u, v);
    ans = s[u] + s[v] - 2 * s[m] + a[m];
    ans += f[u] + f[v] - 2 * f[m] + g[m];
    ans -= h[u] + h[v] - 2 * h[m] - ff[m];
    printf("%lld\n", ans);
  }
  return 0;
}

### 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 std::max;
const int N = 3e5 + 5;
template <typename T>
void read(T &x) {
  bool neg = false;
  unsigned char c = getchar();
  for (; (c ^ 48) > 9; c = getchar())
    if (c == '-') neg = true;
  for (x = 0; (c ^ 48) < 10; c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (neg) x = -x;
}
struct line {
  int next, to, w;
} e[2 * N];
int n, q, hd[N], cnt, a[N], b[N], fa[N], dep[N], siz[N], hc[N], top[N];
long long s[N], f[N], g[N], h[N], c[N], ff[N];
void adline(int u, int v, int w) {
  e[cnt].to = v;
  e[cnt].w = w;
  e[cnt].next = hd[u];
  hd[u] = cnt++;
}
void dfs1(int u) {
  hc[u] = -1;
  s[u] = s[fa[u]] + a[u] - b[u];
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    fa[v] = u;
    b[v] = e[i].w;
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (!~hc[u] || siz[v] > siz[hc[u]]) hc[u] = v;
    f[u] += max(0ll, f[v] + a[v] - 2 * b[v]);
  }
  ++siz[u];
}
void dfs2(int u) {
  long long sum = (ff[u] = f[u]) + g[u] + a[u];
  h[u] = max(h[u], 0ll);
  if (~hc[u]) {
    int v = hc[u];
    top[v] = top[u];
    g[v] = max(0ll, sum - 2 * b[v] - max(0ll, h[v] = f[v] + a[v] - 2 * b[v]));
    dfs2(v);
  }
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u] || v == hc[u]) continue;
    g[v] = max(0ll, sum - 2 * b[v] - max(0ll, h[v] = f[v] + a[v] - 2 * b[v]));
    dfs2(top[v] = v);
  }
}
void dfs3(int u) {
  for (int i = hd[u]; ~i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    f[v] += f[u];
    h[v] += h[u];
    dfs3(v);
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] > dep[top[v]])
      u = fa[top[u]];
    else
      v = fa[top[v]];
  }
  return dep[u] < dep[v] ? u : v;
}
int main() {
  read(n);
  read(q);
  for (int i = 0, i_end = n; i < i_end; ++i) read(a[i]);
  for (int i = 0, i_end = n; i < i_end; ++i) hd[i] = -1;
  for (int i = 0, i_end = n - 1; i < i_end; ++i) {
    int u, v, w;
    read(u);
    read(v);
    read(w);
    adline(--u, --v, w);
    adline(v, u, w);
  }
  dfs1(0);
  dfs2(0);
  dfs3(0);
  while (q--) {
    int u, v, m;
    long long ans;
    read(u);
    read(v);
    --u, --v;
    m = lca(u, v);
    ans = s[u] + s[v] - 2 * s[m] + a[m];
    ans += f[u] + f[v] - 2 * f[m] + g[m];
    ans -= h[u] + h[v] - 2 * h[m] - ff[m];
    printf("%lld\n", ans);
  }
  return 0;
}

```