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. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  x = 0;
  char temp = getchar();
  bool f = false;
  while (!isdigit(temp)) {
    if (temp == '-') f = true;
    temp = getchar();
  }
  while (isdigit(temp)) {
    x = (x << 1) + (x << 3) + temp - '0';
    temp = getchar();
  }
  if (f) x = -x;
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int MAXN = 3e5 + 5;
int n, q;
long long a[MAXN];
struct Edge {
  int to, next;
  long long val;
} edge[MAXN << 1];
int head[MAXN], cnt, from[MAXN];
inline void AddEdge(int u, int v, long long w) {
  edge[++cnt] = (Edge){v, head[u], w}, head[u] = cnt;
}
inline void Add(int u, int v, long long w) {
  AddEdge(u, v, w), AddEdge(v, u, w);
}
int depth[MAXN], ancestor[MAXN][18];
long long fson[MAXN], fbro[MAXN], sumbro[MAXN], ffa[MAXN], sum[MAXN], dis[MAXN];
void Dfs1(int id, int fa) {
  depth[id] = depth[fa] + 1, ancestor[id][0] = fa;
  sum[id] = sum[fa] + a[id];
  for (register int i = 0; i < 17; i++)
    ancestor[id][i + 1] = ancestor[ancestor[id][i]][i];
  for (register int i = head[id]; i; i = edge[i].next) {
    int son = edge[i].to;
    if (son != fa) {
      dis[son] = dis[id] + edge[i].val;
      from[son] = i, Dfs1(son, id);
      fson[id] += max(0ll, fson[son] + a[son] - edge[i].val * 2);
    }
  }
  for (register int i = head[id]; i; i = edge[i].next) {
    int son = edge[i].to;
    if (son != fa)
      fbro[son] = fson[id] - max(fson[son] + a[son] - edge[i].val * 2, 0ll);
  }
}
void Dfs2(int id, int fa) {
  sumbro[id] = sumbro[fa] + fbro[id];
  for (register int i = head[id]; i; i = edge[i].next) {
    int son = edge[i].to;
    if (son != fa) {
      ffa[son] = max(0ll, ffa[id] + a[id] - edge[i].val * 2 + fbro[son]);
      Dfs2(son, id);
    }
  }
}
inline int LCA(int id1, int id2) {
  if (depth[id1] < depth[id2]) swap(id1, id2);
  for (register int i = 17; i >= 0; i--) {
    if (depth[ancestor[id1][i]] >= depth[id2]) id1 = ancestor[id1][i];
    if (id1 == id2) return id1;
  }
  for (register int i = 17; i >= 0; i--)
    if (ancestor[id1][i] != ancestor[id2][i])
      id1 = ancestor[id1][i], id2 = ancestor[id2][i];
  return ancestor[id1][0];
}
int main() {
  read(n), read(q);
  for (register int i = 1; i <= n; i++) read(a[i]);
  for (register int i = 1, u, v, w; i < n; i++)
    read(u), read(v), read(w), Add(u, v, w);
  Dfs1(1, 0), Dfs2(1, 0);
  while (q--) {
    int u, v, lca;
    read(u), read(v), lca = LCA(u, v);
    long long ans = 0;
    ans = sum[u] + sum[v] - sum[lca] - sum[ancestor[lca][0]] -
          (dis[u] - dis[lca]) - (dis[v] - dis[lca]);
    if (depth[u] > depth[v]) swap(u, v);
    if (u == lca)
      ans += ffa[u] + fson[v] + (sumbro[v] - sumbro[u]);
    else {
      ans += ffa[lca] + fson[u] + fson[v] + sumbro[u] + sumbro[v];
      for (register int i = 17; i >= 0; i--) {
        if (depth[ancestor[u][i]] > depth[lca]) u = ancestor[u][i];
        if (depth[ancestor[v][i]] > depth[lca]) v = ancestor[v][i];
      }
      ans -= sumbro[u] + sumbro[v];
      ans += fbro[u] - max(0ll, fson[v] + a[v] - edge[from[v]].val * 2);
    }
    print(ans), puts("");
  }
  return 0;
}