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 <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (1 << 20);
const int MAXLOG = 20;
int n, q;
vector<int> G[MAXN];
map<int, int> we[MAXN];
int value[MAXN];
void read() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> value[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    cin >> we[u][v];
    we[v][u] = we[u][v];
    G[u].push_back(v);
    G[v].push_back(u);
  }
}
int st[MAXN], en[MAXN], dfs_time = 0;
int par[MAXN][MAXLOG];
void dfs_lca(int u) {
  st[u] = ++dfs_time;
  for (int i = 1; i < MAXLOG; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  for (int v : G[u])
    if (v != par[u][0]) {
      par[v][0] = u;
      dfs_lca(v);
    }
  en[u] = dfs_time;
}
inline bool upper(int u, int v) { return st[u] <= st[v] && en[v] <= en[u]; }
int lca(int u, int v) {
  if (upper(u, v)) return u;
  if (upper(v, u)) return v;
  int a = u;
  for (int i = MAXLOG - 1; i >= 0; i--)
    if (!upper(par[a][i], v)) a = par[a][i];
  return par[a][0];
}
int below(int v, int u) {
  int a = u;
  for (int i = MAXLOG - 1; i >= 0; i--)
    if (!upper(par[a][i], v)) a = par[a][i];
  return a;
}
int parent(int u, int up) {
  int a = u;
  for (int i = MAXLOG - 1; i >= 0; i--)
    if (up & (1 << i)) a = par[a][i];
  return a;
}
void lca_precompute(int root) {
  for (int i = 0; i < MAXLOG; i++) par[root][i] = root;
  dfs_time = 0;
  dfs_lca(root);
}
int64_t to_r[MAXN];
int64_t sum[MAXN], dp[MAXN];
void dfs1(int u, int pr) {
  dp[u] = value[u];
  for (int v : G[u])
    if (v != pr) {
      dfs1(v, u);
      chkmax(dp[u], dp[u] + dp[v] - 2ll * we[u][v]);
    }
}
int64_t up[MAXN];
void dfs2(int u, int pr) {
  up[u] = dp[u];
  if (pr != u) {
    int64_t nw = up[u] + up[pr];
    if (dp[u] - 2ll * we[u][pr] >= 0) nw -= dp[u] - 2ll * we[u][pr];
    nw -= 2ll * we[u][pr];
    chkmax(up[u], nw);
  }
  for (int v : G[u])
    if (v != pr) dfs2(v, u);
}
void dfs(int u, int pr) {
  int64_t val = 0;
  if (pr != u) val -= we[u][pr];
  to_r[u] = to_r[pr];
  to_r[u] += dp[u];
  if (pr != u && dp[u] - 2ll * we[u][pr] >= 0) val -= dp[u] - 2ll * we[u][pr];
  to_r[u] += val;
  for (int v : G[u])
    if (v != pr) dfs(v, u);
}
void solve() {
  lca_precompute(1);
  dfs1(1, 1);
  dfs2(1, 1);
  dfs(1, 1);
  while (q--) {
    int u, v, anc;
    cin >> u >> v;
    anc = lca(u, v);
    int64_t answer = up[anc] + to_r[u] + to_r[v] - 2ll * to_r[anc];
    cout << answer << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}