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 maxn = 3e5 + 10;
vector<pair<int, int> > g[maxn];
long long dp[maxn], dp2[maxn];
int n, q;
int a[maxn];
int pa[20][maxn], dep[maxn];
long long upsum[20][maxn];
void dfsin(int x, int fr) {
  dp[x] = a[x];
  pa[0][x] = fr;
  dep[x] = dep[fr] + 1;
  for (int i = 0; i < g[x].size(); i++) {
    int to = g[x][i].first;
    if (to == fr) continue;
    dfsin(to, x);
    dp[x] = max(dp[x], dp[x] + dp[to] - 2ll * g[x][i].second);
  }
}
bool ban[maxn];
int faelen[maxn];
void dfsout(int x, int fr) {
  for (int i = 0; i < g[x].size(); i++) {
    int to = g[x][i].first;
    if (to == fr) continue;
    faelen[to] = g[x][i].second;
    if (dp[to] > 2ll * g[x][i].second) {
      dp2[to] = dp2[x] + dp[x] - a[x] - dp[to] + a[to];
      dp2[to] = max(dp2[to], 1ll * a[to]);
      upsum[0][to] = dp[x] - dp[to] + a[to] + g[x][i].second;
    } else {
      dp2[to] = dp2[x] + dp[x] - a[x] + a[to] - 2ll * g[x][i].second;
      dp2[to] = max(dp2[to], 1ll * a[to]);
      upsum[0][to] = dp[x] + a[to] - 1ll * g[x][i].second;
    }
    dfsout(to, x);
  }
}
void buildlca() {
  for (int k = 1; k < 20; k++) {
    for (int i = 1; i <= n; i++) {
      int fa = pa[k - 1][i];
      pa[k][i] = pa[k - 1][pa[k - 1][i]];
      upsum[k][i] = upsum[k - 1][i] + upsum[k - 1][fa] - a[fa];
    }
  }
}
long long SUM;
int LCA;
void getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  SUM = a[u];
  for (int i = 0; i < 20; i++) {
    if ((dep[u] - dep[v]) >> i & 1) {
      SUM += upsum[i][u] - a[u];
      u = pa[i][u];
    }
  }
  if (u == v) {
    LCA = u;
    return;
  }
  SUM += a[v];
  for (int i = 19; i >= 0; i--) {
    if (pa[i][u] != pa[i][v]) {
      SUM += upsum[i][u] - a[u];
      SUM += upsum[i][v] - a[v];
      u = pa[i][u];
      v = pa[i][v];
    }
  }
  LCA = pa[0][u];
  SUM += upsum[0][u] - a[u];
  if (dp[v] > 2ll * faelen[v]) {
    SUM -= dp[v];
    SUM += faelen[v];
  } else {
    SUM -= faelen[v];
  }
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    g[u].push_back(make_pair(v, w));
    g[v].push_back(make_pair(u, w));
  }
  dfsin(1, 0);
  dp2[1] = a[1];
  upsum[0][1] = a[1];
  dfsout(1, 0);
  buildlca();
  while (q--) {
    int u, v;
    scanf("%d%d", &u, &v);
    if (u == v) {
      printf("%I64d\n", dp[u] + dp2[u] - a[u]);
      continue;
    }
    getlca(u, v);
    if (LCA != u && LCA != v) {
      printf("%I64d\n", SUM + dp[u] + dp[v] - a[u] - a[v] + dp2[LCA] - a[LCA]);
    } else if (LCA == u) {
      printf("%I64d\n", SUM + dp[v] - a[v] + dp2[u] - a[u]);
    } else {
      printf("%I64d\n", SUM + dp[u] - a[u] + dp2[v] - a[v]);
    }
  }
}