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>
template <class T>
bool chmax(T &val, T nv) {
  return val < nv ? (val = nv, true) : false;
}
template <class T>
bool chmin(T &val, T nv) {
  return nv < val ? (val = nv, true) : false;
}
using namespace std;
using ll = long long;
const ll MAX_N = 300010, MAX_K = 20;
ll n, q, a[MAX_N];
ll pf[MAX_N], dn[MAX_N], gpf[MAX_N], up[MAX_N], dp[MAX_N];
ll anc[MAX_K][MAX_N], in[MAX_N], out[MAX_N];
vector<pair<ll, ll>> edge[MAX_N];
bool isanc(ll a, ll b) { return in[a] <= in[b] && out[a] >= out[b]; }
ll getlca(ll a, ll b) {
  for (ll d = MAX_K - 1; d >= 0; --d)
    if (!isanc(anc[d][a], b)) a = anc[d][a];
  return isanc(a, b) ? a : anc[0][a];
}
void dfs1(ll x, ll lst) {
  static ll t;
  anc[0][x] = lst;
  in[x] = t++;
  for (auto [u, w] : edge[x])
    if (u != lst) dfs1(u, x);
  out[x] = t;
}
void dfs2(ll x, ll lst) {
  pf[x] += a[x];
  dp[x] = a[x];
  for (auto [u, w] : edge[x])
    if (u != lst) {
      pf[u] = -w;
      dfs2(u, x);
      dp[x] += max(0ll, dp[u] - 2 * w);
    }
}
void dfs3(ll x, ll lst) {
  for (auto [u, w] : edge[x])
    if (u != lst) {
      up[u] = max(0ll, up[x] + dp[x] - max(0ll, dp[u] - 2 * w) - 2 * w);
      gpf[u] = max(0ll, dp[u] - w * 2);
      dfs3(u, x);
    }
}
void dfs4(ll x, ll lst) {
  pf[x] += dp[x] - a[x];
  for (auto [u, w] : edge[x])
    if (u != lst) {
      pf[u] += pf[x], gpf[u] += gpf[x];
      dfs4(u, x);
    }
}
void proc() {
  dfs1(1, 1);
  for (ll d = 1; d < MAX_K; ++d)
    for (ll i = 1; i <= n; ++i) anc[d][i] = anc[d - 1][anc[d - 1][i]];
  dfs2(1, 1);
  dfs3(1, 1);
  dfs4(1, 1);
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  for (ll i = 1; i <= n; ++i) cin >> a[i];
  for (ll a, b, w, i = 1; i < n; ++i) {
    cin >> a >> b >> w;
    edge[a].emplace_back(b, w);
    edge[b].emplace_back(a, w);
  }
  proc();
  for (ll i = 1; i <= n; ++i) 0;
  while (q--) {
    ll a, b, lca;
    cin >> a >> b;
    lca = getlca(a, b);
    0;
    0;
    cout << (pf[a] + pf[b] - 2 * pf[lca] + dp[lca] + up[lca]) -
                (gpf[a] + gpf[b] - 2 * gpf[lca])
         << '\n';
  }
}