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>
const uint32_t INF = 1e9;
struct Vertex;
struct Edge {
  Vertex *to;
  int64_t w;
  Edge(Vertex *to = nullptr, int64_t w = 0) : to(to), w(w) {}
  friend bool operator==(const Edge &a, const Edge &b) {
    return (a.to == b.to) && (a.w == b.w);
  }
};
struct Vertex {
  std::vector<Edge> chld;
  Edge par;
  uint32_t a;
  uint32_t depth;
  std::vector<Edge> skips;
  int64_t down, up;
  Vertex() : a(0), depth(0), down(0), up(0) {}
  void root_tree() {
    for (auto it = chld.begin(); it != chld.end(); it++) {
      if (*it == par) {
        chld.erase(it);
        break;
      }
    }
    for (const Edge &e : chld) {
      e.to->par = Edge(this, e.w);
      e.to->depth = depth + 1;
      e.to->root_tree();
    }
  }
  void calc_down() {
    down = a;
    for (const Edge &e : chld) {
      e.to->calc_down();
      if (e.to->down > 2 * e.w) {
        down += e.to->down;
        down -= 2 * e.w;
      }
    }
  }
  void calc_up() {
    up = 0;
    if (par.to != nullptr) {
      up = par.to->up + par.to->down;
      if (down > 2 * par.w) {
        up += 2 * par.w;
        up -= down;
      }
      if (up > 2 * par.w) {
        up -= 2 * par.w;
      } else {
        up = 0;
      }
    }
    for (const Edge &e : chld) {
      e.to->calc_up();
    }
  }
  void build_skips() {
    if (par.to != nullptr) {
      int64_t curDown = par.to->down;
      if (down > 2 * par.w) {
        curDown += 2 * par.w;
        curDown -= down;
      }
      curDown -= par.w;
      skips.push_back({par.to, curDown});
      for (uint32_t i = 0; i < skips[i].to->skips.size(); i++) {
        skips.push_back(
            {skips[i].to->skips[i].to, skips[i].w + skips[i].to->skips[i].w});
      }
    }
    for (const Edge &e : chld) {
      e.to->build_skips();
    }
  }
  static int64_t calc_ans(Vertex *a, Vertex *b) {
    if (b->depth > a->depth) {
      std::swap(a, b);
    }
    if (a == b) {
      return a->down + a->up;
    }
    int64_t pathW = a->down;
    for (uint32_t i = a->skips.size() - 1; i < INF; i--) {
      if (a->skips.size() <= i) {
        continue;
      }
      if (a->skips[i].to->depth >= b->depth) {
        pathW += a->skips[i].w;
        a = a->skips[i].to;
      }
    }
    if (a == b) {
      return pathW + a->up;
    }
    pathW += b->down;
    for (uint32_t i = a->skips.size() - 1; i < INF; i--) {
      if (a->skips.size() <= i) {
        continue;
      }
      if (a->skips[i].to != b->skips[i].to) {
        pathW += a->skips[i].w;
        pathW += b->skips[i].w;
        a = a->skips[i].to;
        b = b->skips[i].to;
      }
    }
    pathW += a->par.to->down;
    pathW += a->par.to->up;
    pathW -= a->par.w;
    if (a->down > 2 * a->par.w) {
      pathW += 2 * a->par.w;
      pathW -= a->down;
    }
    pathW -= b->par.w;
    if (b->down > 2 * b->par.w) {
      pathW += 2 * b->par.w;
      pathW -= b->down;
    }
    return pathW;
  }
};
std::vector<Vertex> tree;
uint32_t qCnt;
void input() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  uint32_t vCnt;
  std::cin >> vCnt >> qCnt;
  tree.resize(vCnt);
  for (Vertex &v : tree) {
    std::cin >> v.a;
  }
  for (uint32_t i = 1; i < vCnt; i++) {
    uint32_t u, v, w;
    std::cin >> u >> v >> w;
    u--;
    v--;
    tree[u].chld.push_back({&tree[v], w});
    tree[v].chld.push_back({&tree[u], w});
  }
}
int main() {
  input();
  tree[0].root_tree();
  tree[0].calc_down();
  tree[0].calc_up();
  tree[0].build_skips();
  for (uint32_t i = 0; i < qCnt; i++) {
    uint32_t a, b;
    std::cin >> a >> b;
    std::cout << Vertex::calc_ans(&tree[a - 1], &tree[b - 1]) << "\n";
  }
}