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 N = 4e5 + 5, LOG = 20;
vector<pair<int, long long> > G[N];
vector<int> Query[N];
int Timer;
int st[N];
int en[N];
long long val[N];
long long dp[N];
long long cont[N];
long long cum[N];
long long tree[4 * N];
long long lazy[4 * N];
long long res[N];
int par[N][LOG];
int depth[N];
int from[N];
int to[N];
int n, Q;
void dfs1(int u, int p) {
  st[u] = ++Timer;
  dp[u] = val[u];
  for (auto pp : G[u]) {
    int v = pp.first;
    long long w = pp.second;
    if (v != p) {
      dfs1(v, u);
      long long cnt = max(0LL, dp[v] - 2 * w);
      cont[v] = cnt;
      dp[u] += cnt;
    }
  }
  en[u] = Timer;
}
void dfs2(int u, int p, long long w) {
  cum[u] += cum[p] + dp[u] - cont[u] - w;
  for (auto pp : G[u]) {
    int v = pp.first;
    long long w = pp.second;
    if (v != p) {
      dfs2(v, u, w);
    }
  }
}
void relax(int cn, int b, int e) {
  if (lazy[cn]) {
    tree[cn] += lazy[cn] * (e - b + 1);
    if (b != e) {
      lazy[2 * cn] += lazy[cn];
      lazy[2 * cn + 1] += lazy[cn];
    }
    lazy[cn] = 0;
  }
}
void upd(int cn, int b, int e, int i, int j, long long add) {
  relax(cn, b, e);
  if (b > j or e < i) {
    return;
  }
  int lc = 2 * cn, rc = lc + 1, mid = (b + e) / 2;
  if (b >= i and e <= j) {
    lazy[cn] = add;
    relax(cn, b, e);
    return;
  }
  upd(lc, b, mid, i, j, add);
  upd(rc, mid + 1, e, i, j, add);
  tree[cn] = tree[lc] + tree[rc];
}
long long query(int cn, int b, int e, int i, int j) {
  relax(cn, b, e);
  if (b > j or e < i) return 0;
  if (b >= i and e <= j) {
    return tree[cn];
  }
  int lc = 2 * cn, rc = lc + 1, mid = (b + e) / 2;
  return query(lc, b, mid, i, j) + query(rc, mid + 1, e, i, j);
}
void dfs3(int u, int p, int lvl) {
  par[u][0] = p;
  depth[u] = lvl;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i].first;
    if (v == p) continue;
    dfs3(v, u, lvl + 1);
  }
}
void init(int root, int n) {
  dfs3(root, -1, 1);
  for (int k = 1; k < LOG; k++) {
    for (int i = 1; i <= n; i++) {
      if (par[i][k - 1] != -1)
        par[i][k] = par[par[i][k - 1]][k - 1];
      else
        par[i][k] = -1;
    }
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) {
    swap(u, v);
  }
  int diff = depth[u] - depth[v];
  for (int i = LOG - 1; i >= 0; i--) {
    if (diff >= (1 << i)) {
      diff -= (1 << i);
      u = par[u][i];
    }
  }
  if (u == v) return u;
  for (int i = LOG - 1; i >= 0; i--) {
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  }
  return par[v][0];
}
void solve(int root) {
  for (auto q : Query[root]) {
    int u = from[q], v = to[q];
    res[q] = query(1, 1, n, st[u], st[u]) + query(1, 1, n, st[v], st[v]) -
             2 * query(1, 1, n, st[root], st[root]) + dp[root];
  }
}
void dfs4(int u, int p, long long weight) {
  long long dpu = dp[u], dpp = dp[p], contu = cont[u], contp = cont[p];
  if (p) {
    dp[p] = dp[p] - cont[u];
    long long dpAdd = max(0LL, dp[p] - 2 * weight);
    cont[p] = dpAdd;
    cont[u] = 0;
    dp[u] += dpAdd;
    upd(1, 1, n, st[u], en[u], dp[u] - dpu - cont[u] + contu);
    upd(1, 1, n, st[p], en[p], dp[p] - dpp - cont[p] + contp);
  }
  solve(u);
  for (auto pp : G[u]) {
    int v = pp.first;
    long long w = pp.second;
    if (v != p) {
      dfs4(v, u, w);
    }
  }
  if (p) {
    upd(1, 1, n, st[u], en[u], -(dp[u] - dpu - cont[u] + contu));
    upd(1, 1, n, st[p], en[p], -(dp[p] - dpp - cont[p] + contp));
    dp[u] = dpu;
    dp[p] = dpp;
    cont[u] = contu, cont[p] = contp;
  }
}
int main() {
  cin >> n >> Q;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &val[i]);
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    long long w;
    scanf("%d %d %lld", &u, &v, &w);
    G[u].push_back({v, w});
    G[v].push_back({u, w});
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; i++) {
    upd(1, 1, n, st[i], st[i], cum[i]);
  }
  init(1, n);
  for (int i = 1; i <= Q; i++) {
    scanf("%d %d", &from[i], &to[i]);
    int Lca = lca(from[i], to[i]);
    Query[Lca].push_back(i);
  }
  dfs4(1, -1, 0);
  for (int i = 1; i <= Q; i++) {
    printf("%lld\n", res[i]);
  }
}