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 + 100;
vector<int> edge[N];
vector<long long> w[N];
int n, q, top[N], dfn[N], hson[N], siz[N];
int fa[N];
long long val[N], f[N], g[N];
long long a[N], dv[N];
long long dep[N], faw[N];
void dfs(int x) {
  f[x] = 0;
  siz[x] = 1;
  dv[x] = dv[fa[x]] + a[x];
  for (int i = 0; i <= (int)edge[x].size() - 1; ++i) {
    int y = edge[x][i];
    if (y == fa[x]) continue;
    fa[y] = x;
    dep[y] = dep[x] + w[x][i];
    faw[y] = w[x][i];
    dfs(y);
    siz[x] += siz[y];
    if (siz[y] > siz[hson[x]]) hson[x] = y;
    f[x] = max(f[x], f[x] + f[y] - w[x][i] * 2 + a[y]);
  }
  for (int i = 0; i <= (int)edge[x].size() - 1; ++i) {
    int y = edge[x][i];
    if (y == fa[x]) continue;
    if (f[y] - w[x][i] * 2 + a[y] > 0)
      val[y] = f[x] - f[y] + w[x][i] * 2 - a[y];
    else
      val[y] = f[x];
  }
}
int tot, an[N];
void dfs2(int x, int k) {
  top[x] = k;
  dfn[x] = ++tot;
  an[tot] = x;
  if (!hson[x]) return;
  dfs2(hson[x], k);
  for (int i = 0; i <= (int)edge[x].size() - 1; ++i) {
    int y = edge[x][i];
    if (y == fa[x] || y == hson[x]) continue;
    dfs2(y, y);
  }
}
void dfs3(int x) {
  for (int i = 0; i <= (int)edge[x].size() - 1; ++i) {
    int y = edge[x][i];
    if (y == fa[x]) continue;
    g[y] = g[x] + a[x] - 2 * w[x][i];
    if (f[y] - w[x][i] * 2 + a[y] < 0)
      g[y] += f[x];
    else
      g[y] += f[x] - f[y] + w[x][i] * 2 - a[y];
    if (g[y] < 0) g[y] = 0;
    dfs3(y);
  }
}
inline int lca(int x, int y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  return (dep[x] < dep[y]) ? x : y;
}
long long st[N];
void init() {
  st[0] = 0;
  for (int i = 1; i <= n; ++i) st[i] = st[i - 1] + val[an[i]];
}
inline long long rmq(int l, int r) { return st[r] - st[l - 1]; }
int son;
long long query(int x, int y) {
  long long res = 0;
  while (top[x] != top[y]) {
    res += rmq(dfn[top[x]], dfn[x]);
    if (fa[top[x]] == y) res -= val[top[x]], son = top[x];
    x = fa[top[x]];
  }
  if (!son) son = an[dfn[y] + 1];
  if (dfn[y] + 2 <= dfn[x]) res += rmq(dfn[y] + 2, dfn[x]);
  return res;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) scanf("%lld", a + i);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    long long z;
    scanf("%d%d%lld", &u, &v, &z);
    edge[u].push_back(v);
    edge[v].push_back(u);
    w[u].push_back(z);
    w[v].push_back(z);
  }
  fa[1] = dep[1] = 0;
  dfs(1);
  dfs2(1, 1);
  dfs3(1);
  init();
  for (int i = 1; i <= q; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    int w = lca(x, y);
    long long res = 0;
    son = 0;
    if (w == x) {
      if (w == y)
        res = f[w] + g[w];
      else {
        res = query(y, w);
        res += val[son] + g[w] + f[y];
      }
    } else if (w == y) {
      res = query(x, w);
      res += val[son] + g[w] + f[x];
    } else {
      res = query(x, w) + g[w];
      if (f[son] - 2 * faw[son] + a[son] > 0)
        res -= f[son] - 2 * faw[son] + a[son];
      son = 0;
      res += query(y, w);
      res += val[son];
      res += f[x] + f[y];
    }
    res += dv[x] + dv[y] - dv[w] - dv[fa[w]] - dep[x] - dep[y] + dep[w] * 2;
    printf("%lld\n", res);
  }
  return 0;
}