#include <bits/stdc++.h>
using namespace std;
int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x;
}
const int N = 4e5 + 10;
struct Edge {
  int nxt, point, w;
} edge[N << 1];
int head[N], tot, n, a[N];
long long f[N], g[N];
int wson[N], siz[N], dep[N], top[N], fa[N], dfn[N], dfntot;
long long disa[N], disw[N];
long long Df[N], DF[N];
void add_edge(int u, int v, int w) {
  edge[++tot].nxt = head[u];
  edge[tot].point = v;
  edge[tot].w = w;
  head[u] = tot;
}
void dfs(int u) {
  f[u] = a[u];
  siz[u] = 1;
  dep[u] = dep[fa[u]] + 1;
  disa[u] += a[u];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa[u]) continue;
    fa[v] = u;
    disa[v] = disa[u];
    disw[v] = disw[u] + edge[i].w;
    dfs(v);
    siz[u] += siz[v];
    if (siz[wson[u]] < siz[v]) wson[u] = v;
    f[u] += max(0ll, f[v] - 2 * edge[i].w);
  }
}
void treedp(int u, int fa) {
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa) continue;
    f[u] -= max(0ll, f[v] - 2 * edge[i].w);
    g[v] = max(0ll, f[u] - 2 * edge[i].w);
    f[v] += g[v];
    treedp(v, u);
    f[v] -= g[v];
    f[u] += max(0ll, f[v] - 2 * edge[i].w);
  }
}
long long F[N];
void dfs2(int u, int topver) {
  dfn[u] = ++dfntot;
  top[u] = topver;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (!dfn[v]) f[v] = max(f[v] - 2 * edge[i].w, 0ll), F[u] += f[v];
  }
  if (wson[u])
    dfs2(wson[u], topver);
  else
    return;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (!dfn[v]) {
      dfs2(v, v);
    }
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    u = fa[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  return u;
}
void dfs3(int u) {
  Df[u] += f[u], DF[u] += F[u];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa[u]) continue;
    Df[v] = Df[u], DF[v] = DF[u];
    dfs3(v);
  }
}
void solve(int u, int v) {
  int x = lca(u, v);
  long long ans = (disa[u] + disa[v] - 2 * disa[x] + a[x]) -
                  (disw[u] + disw[v] - 2 * disw[x]);
  ans += g[x];
  ans += (DF[u] + DF[v] - 2 * DF[x] + F[x]) - (Df[u] + Df[v] - 2 * Df[x]);
  printf("%lld\n", ans);
}
int main() {
  n = read();
  int q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    add_edge(u, v, w);
    add_edge(v, u, w);
  }
  dfs(1);
  treedp(1, 0);
  dfs2(1, 1);
  dfs3(1);
  while (q--) {
    int u = read(), v = read();
    solve(u, v);
  }
  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 namespace std;
int read() {
  char c = getchar();
  int x = 0;
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x;
}
const int N = 4e5 + 10;
struct Edge {
  int nxt, point, w;
} edge[N << 1];
int head[N], tot, n, a[N];
long long f[N], g[N];
int wson[N], siz[N], dep[N], top[N], fa[N], dfn[N], dfntot;
long long disa[N], disw[N];
long long Df[N], DF[N];
void add_edge(int u, int v, int w) {
  edge[++tot].nxt = head[u];
  edge[tot].point = v;
  edge[tot].w = w;
  head[u] = tot;
}
void dfs(int u) {
  f[u] = a[u];
  siz[u] = 1;
  dep[u] = dep[fa[u]] + 1;
  disa[u] += a[u];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa[u]) continue;
    fa[v] = u;
    disa[v] = disa[u];
    disw[v] = disw[u] + edge[i].w;
    dfs(v);
    siz[u] += siz[v];
    if (siz[wson[u]] < siz[v]) wson[u] = v;
    f[u] += max(0ll, f[v] - 2 * edge[i].w);
  }
}
void treedp(int u, int fa) {
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa) continue;
    f[u] -= max(0ll, f[v] - 2 * edge[i].w);
    g[v] = max(0ll, f[u] - 2 * edge[i].w);
    f[v] += g[v];
    treedp(v, u);
    f[v] -= g[v];
    f[u] += max(0ll, f[v] - 2 * edge[i].w);
  }
}
long long F[N];
void dfs2(int u, int topver) {
  dfn[u] = ++dfntot;
  top[u] = topver;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (!dfn[v]) f[v] = max(f[v] - 2 * edge[i].w, 0ll), F[u] += f[v];
  }
  if (wson[u])
    dfs2(wson[u], topver);
  else
    return;
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (!dfn[v]) {
      dfs2(v, v);
    }
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    u = fa[top[u]];
  }
  if (dep[u] > dep[v]) swap(u, v);
  return u;
}
void dfs3(int u) {
  Df[u] += f[u], DF[u] += F[u];
  for (int i = head[u]; i; i = edge[i].nxt) {
    int v = edge[i].point;
    if (v == fa[u]) continue;
    Df[v] = Df[u], DF[v] = DF[u];
    dfs3(v);
  }
}
void solve(int u, int v) {
  int x = lca(u, v);
  long long ans = (disa[u] + disa[v] - 2 * disa[x] + a[x]) -
                  (disw[u] + disw[v] - 2 * disw[x]);
  ans += g[x];
  ans += (DF[u] + DF[v] - 2 * DF[x] + F[x]) - (Df[u] + Df[v] - 2 * Df[x]);
  printf("%lld\n", ans);
}
int main() {
  n = read();
  int q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    add_edge(u, v, w);
    add_edge(v, u, w);
  }
  dfs(1);
  treedp(1, 0);
  dfs2(1, 1);
  dfs3(1);
  while (q--) {
    int u = read(), v = read();
    solve(u, v);
  }
  return 0;
}

```