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;
const int MAX_N = 300000;
const int MAX_LOG_N = 19;
struct Edge {
  int src;
  int dest;
  int weight;
  Edge(int s = 0, int d = 0, int w = 0) : src(s), dest(d), weight(w) {}
};
int N, Q;
int V[MAX_N];
Edge E[2 * MAX_N];
vector<int> edges[MAX_N];
int visitCount;
int visit[MAX_N], leave[MAX_N];
long long int subTreeCycles[MAX_N];
long long int edgeCycle[2 * MAX_N];
long long int vertexCycle[MAX_N];
long long int ancestor[MAX_LOG_N][MAX_N];
long long int twoPath[MAX_LOG_N][MAX_N];
int upEdge[MAX_LOG_N][MAX_N];
int downEdge[MAX_LOG_N][MAX_N];
void inputTree() {
  cin >> N >> Q;
  for (int i = 0; i < N; ++i) {
    cin >> V[i];
  }
  for (int i = 0; i < N - 1; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    E[i * 2] = Edge(u, v, w);
    edges[u].push_back(i * 2);
    E[i * 2 + 1] = Edge(v, u, w);
    edges[v].push_back(i * 2 + 1);
  }
}
void computeSubTreeCycles(int u, int d) {
  visit[u] = ++visitCount;
  for (int idx : edges[u]) {
    int v = E[idx].dest;
    if (!visit[v]) {
      computeSubTreeCycles(v, d + 1);
      edgeCycle[idx] = max(0LL, subTreeCycles[v] + V[v] - 2 * E[idx].weight);
      subTreeCycles[u] += edgeCycle[idx];
    }
  }
  leave[u] = ++visitCount;
}
void dump() {
  cout << "Vertices" << endl;
  for (int i = 0; i < N; ++i) {
    cout << i + 1 << ": " << subTreeCycles[i] << ' ' << vertexCycle[i] << endl;
  }
  cout << "Edges" << endl;
  for (int i = 0; i < 2 * (N - 1); ++i) {
    cout << '(' << E[i].src + 1 << ',' << E[i].dest + 1 << ")->" << E[i].weight
         << ": " << edgeCycle[i] << endl;
  }
}
void computeVertexCycle(int u) {
  for (int idx : edges[u]) {
    int v = E[idx].dest;
    if (visit[v] > visit[u]) {
      int reverseIdx = (idx & 1) ? idx - 1 : idx + 1;
      edgeCycle[reverseIdx] =
          max(0LL, vertexCycle[u] - edgeCycle[idx] - 2 * E[idx].weight);
      vertexCycle[v] = subTreeCycles[v] + edgeCycle[reverseIdx] + V[v];
      computeVertexCycle(v);
      ancestor[0][v] = u;
      twoPath[0][v] = vertexCycle[v] - edgeCycle[idx] - edgeCycle[reverseIdx] -
                      E[idx].weight;
      upEdge[0][v] = reverseIdx, downEdge[0][v] = idx;
    }
  }
}
void computeLca() {
  for (int i = 0; i < MAX_LOG_N; ++i) {
    ancestor[i][0] = -1;
  }
  for (int k = 1; k < MAX_LOG_N; ++k) {
    for (int v = 1; v < N; ++v) {
      int u = ancestor[k - 1][v];
      if (u == -1) {
        ancestor[k][v] = -1;
        continue;
      }
      ancestor[k][v] = ancestor[k - 1][u];
      if (ancestor[k][v] != -1) {
        upEdge[k][v] = upEdge[k - 1][v];
        downEdge[k][v] = downEdge[k - 1][u];
        twoPath[k][v] = twoPath[k - 1][v] + twoPath[k - 1][u];
      }
    }
  }
}
void dumpLca() {
  for (int v = 0; v < N; ++v) {
    cout << v + 1 << ':';
    for (int i = 0; i < MAX_LOG_N; ++i) {
      cout << ' ' << ancestor[i][v] + 1 << '(' << twoPath[i][v] << ')';
    }
    cout << endl;
  }
}
bool isParent(int u, int v) {
  return visit[u] <= visit[v] && leave[u] >= visit[v];
}
long long int answer(int u, int v) {
  long long int sum = 0LL;
  for (int i = MAX_LOG_N - 1; i >= 0; --i) {
    if (ancestor[i][u] != -1 && !isParent(ancestor[i][u], v)) {
      sum += twoPath[i][u];
      u = ancestor[i][u];
    }
  }
  if (!isParent(u, v)) {
    sum += twoPath[0][u];
    u = ancestor[0][u];
  }
  for (int i = MAX_LOG_N - 1; i >= 0; --i) {
    if (ancestor[i][v] != -1) {
      if (ancestor[i][v] == u || !isParent(ancestor[i][v], u)) {
        sum += twoPath[i][v];
        v = ancestor[i][v];
      }
    }
  }
  return sum + vertexCycle[u];
}
void answerQueries() {
  int u, v;
  for (int i = 0; i < Q; ++i) {
    cin >> u >> v;
    --u, --v;
    cout << answer(u, v) << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  inputTree();
  computeSubTreeCycles(0, 0);
  vertexCycle[0] = subTreeCycles[0] + V[0];
  computeVertexCycle(0);
  computeLca();
  answerQueries();
  return 0;
}