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 = 3e5 + 10;
vector<int> v[N], edge[N];
void add(int x, int y, int z) {
  v[x].push_back(y);
  edge[x].push_back(z);
}
long long fs[N], fe[N], gout[N], sfs[N], sfe[N], sw[N];
int dep[N], sz[N], son[N], pre[N], top[N], fro[N], a[N];
int n, q, in_li[N], in_tr[N], tot;
void dfs1(int x, int fa) {
  pre[x] = fa;
  sz[x] = 1;
  dep[x] = dep[fa] + 1;
  int l = v[x].size(), y;
  fs[x] = a[x];
  for (register int i = (0); i <= (l - 1); ++i) {
    y = v[x][i];
    if (y == fa) continue;
    dfs1(y, x);
    fro[y] = edge[x][i];
    fs[x] += max(0ll, fs[y] - 2 * fro[y]);
    sz[x] += sz[y];
    if (sz[son[x]] < sz[y]) son[x] = y;
  }
}
void dfs2(int x) {
  if (x == 1)
    gout[x] = a[x];
  else {
    long long tp =
        gout[pre[x]] + fs[pre[x]] - max(0ll, fs[x] - 2 * fro[x]) - a[pre[x]];
    gout[x] = a[x] + max(0ll, tp - 2 * fro[x]);
    fe[x] = fs[pre[x]] - max(0ll, fs[x] - 2 * fro[x]);
  }
  sfs[x] = sfs[pre[x]] + fs[x];
  sfe[x] = sfe[pre[x]] + fe[x];
  sw[x] = sw[pre[x]] + fro[x];
  int l = v[x].size(), y;
  for (register int i = (0); i <= (l - 1); ++i) {
    y = v[x][i];
    if (y == pre[x]) continue;
    dfs2(y);
  }
}
void dfs3(int x) {
  if (son[x]) {
    in_li[son[x]] = ++tot;
    in_tr[tot] = son[x];
    top[son[x]] = top[x];
    dfs3(son[x]);
  }
  int l = v[x].size(), y;
  for (register int i = (0); i <= (l - 1); ++i) {
    y = v[x][i];
    if (y == pre[x] || y == son[x]) continue;
    in_li[y] = ++tot;
    in_tr[tot] = y;
    top[y] = y;
    dfs3(y);
  }
}
int lca(int x, int y) {
  int fx = top[x], fy = top[y];
  while (fx != fy) {
    if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
    x = pre[fx];
    fx = top[x];
  }
  if (dep[x] > dep[y]) swap(x, y);
  return x;
}
int jump(int x, int z) {
  if (x == z) return -1;
  int fx = top[x];
  while (fx != top[z]) {
    if (pre[fx] == z) return fx;
    x = pre[fx];
    fx = top[x];
  }
  return in_tr[in_li[z] + 1];
}
long long query(int x, int y) {
  int z = lca(x, y), fx = jump(x, z), fy = jump(y, z);
  long long ans = gout[z] + fs[z] - a[z];
  if (x != z) {
    ans += fs[x] + sfe[x] - sfe[fx];
    ans -= (sw[x] - sw[z]);
    ans -= max(0ll, fs[fx] - 2 * fro[fx]);
  }
  if (y != z) {
    ans += fs[y] + sfe[y] - sfe[fy];
    ans -= (sw[y] - sw[z]);
    ans -= max(0ll, fs[fy] - 2 * fro[fy]);
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int x, y, z;
  cin >> n >> q;
  for (register int i = (1); i <= (n); ++i) cin >> a[i];
  for (register int i = (1); i <= (n - 1); ++i) {
    cin >> x >> y >> z;
    add(x, y, z);
    add(y, x, z);
  }
  dfs1(1, 0);
  dfs2(1);
  top[1] = in_li[1] = in_tr[1] = tot = 1;
  dfs3(1);
  while (q--) cin >> x >> y, cout << query(x, y) << '\n';
  return 0;
}