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 INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 1e9 + 7;
const long long maxn = 3e5 + 7;
const double pi = acos(-1.0);
const double eps = 0.00000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <typename T>
inline T powMM(T a, T b) {
  T ret = 1;
  for (; b; b >>= 1ll, a = (long long)a * a % M)
    if (b & 1) ret = (long long)ret * a % M;
  return ret;
}
int A[maxn];
vector<pair<int, int> > edge[maxn];
int dep[maxn];
long long len[maxn], dp_add[maxn], dp[maxn], dp_down[maxn];
int fa[maxn][20];
void dfs_up(int x, int father, int depth, long long length) {
  len[x] = length;
  dep[x] = depth;
  dp[x] = 0;
  int i;
  fa[x][0] = father;
  for (i = 1; i < 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (auto now : edge[x])
    if (now.first != father) {
      int v = now.first;
      dfs_up(v, x, depth + 1, length + now.second);
      dp_add[v] = max(0ll, dp[v] - now.second * 2 + A[v]);
      dp[x] += dp_add[v];
    }
}
long long sum_dp[maxn], sum_dp_add[maxn], sum_A[maxn];
void dfs_down(int x, int fa) {
  sum_A[x] = A[x] + sum_A[fa];
  sum_dp[x] = dp[x] + sum_dp[fa];
  sum_dp_add[x] = dp_add[x] + sum_dp_add[fa];
  for (auto now : edge[x])
    if (now.first != fa) {
      int v = now.first;
      dp_down[v] =
          max(0ll, dp_down[x] + A[x] + dp[x] - dp_add[v] - now.second * 2);
      dfs_down(v, x);
    }
}
inline int lca(int x, int y) {
  int i;
  if (dep[x] < dep[y]) swap(x, y);
  for (i = 20 - 1; i >= 0; i--)
    if (dep[x] - dep[y] >= 1 << i) x = fa[x][i];
  if (x == y) return x;
  for (i = 20 - 1; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
char buffer[36000000], *buf = buffer;
void read(int &x) {
  for (x = 0; *buf < 48; ++buf)
    ;
  while (*buf >= 48) x = x * 10 + *buf - 48, ++buf;
}
int TaskA() {
  int n, q, i;
  fread(buffer, 1, 36000000, stdin);
  read(n);
  read(q);
  for (i = 1; i <= n; i++) read(A[i]);
  for (i = 1; i <= n - 1; i++) {
    int u, v, len;
    read(u);
    read(v);
    read(len);
    edge[u].push_back(make_pair(v, len));
    edge[v].push_back(make_pair(u, len));
  }
  dfs_up(1, 0, 0, 0);
  dfs_down(1, 0);
  for (i = 1; i <= q; i++) {
    int u, v;
    read(u);
    read(v);
    int f = lca(u, v);
    long long ans = 0;
    ans = sum_A[v] + sum_A[u] - sum_A[f] - sum_A[fa[f][0]];
    ans -= len[v] + len[u] - len[f] * 2;
    ans += sum_dp[v] + sum_dp[u] - sum_dp[f] - sum_dp[fa[f][0]];
    ans -= sum_dp_add[v] + sum_dp_add[u] - sum_dp_add[f] * 2;
    ans += dp_down[f];
    printf("%I64d\n", ans);
  }
  return 0;
}
int main() {
  int startTime = clock();
  fprintf(stderr, "/--- initializeTime: %ld milliseconds ---/\n",
          clock() - startTime);
  int T = 1;
  startTime = clock();
  while (T--) TaskA();
  fprintf(stderr, "/--- computeTime: %ld milliseconds ---/\n",
          clock() - startTime);
}