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;
template <typename T>
bool chkmax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename T>
bool chkmin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename T>
T getSign(const T &a) {
  return (a > T(0)) - (a < T(0));
}
const double pi = acos(-1);
void procStatus() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
int read() {
  int x = 0, flag = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flag *= -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  return x * flag;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
const int Maxn = 3e5 + 9, Maxq = 4e5 + 9;
struct edge {
  int to, nxt;
  long long w;
} g[Maxn << 1];
int head[Maxn], e;
void add(int u, int v, int w) { g[++e] = (edge){v, head[u], w}, head[u] = e; }
int n, q, a[Maxn];
void Init() {
  memset((head), (-1), sizeof(head));
  n = read(), q = read();
  for (int i = (1), i_end_ = (n); i <= i_end_; ++i) a[i] = read();
  for (int i = (1), i_end_ = (n - 1); i <= i_end_; ++i) {
    int u = read(), v = read(), w = read();
    add(u, v, w), add(v, u, w);
  }
}
int fa[21][Maxn], dep[Maxn];
long long tov[Maxn], dis[Maxn], valDis[Maxn];
void dfsFa(int u, int pa) {
  fa[0][u] = pa;
  dep[u] = dep[pa] + 1, valDis[u] = valDis[pa] + a[u];
  for (int i = (1), i_end_ = (20); i <= i_end_; ++i)
    fa[i][u] = fa[i - 1][fa[i - 1][u]];
  for (int i = head[u]; ~i; i = g[i].nxt) {
    int v = g[i].to;
    if (v != pa) {
      dis[v] = dis[u] + g[i].w;
      tov[v] = g[i].w, dfsFa(v, u);
    }
  }
}
long long dp[Maxn][4], used[Maxn], suffix[Maxn];
void dfsDp12(int u, int pa) {
  for (int i = head[u]; ~i; i = g[i].nxt) {
    int v = g[i].to;
    if (v != pa) {
      dfsDp12(v, u);
      if (dp[v][1] + a[v] - 2 * g[i].w >= 0)
        dp[u][1] += dp[v][1] + a[v] - 2ll * g[i].w, used[v] = 1;
    }
  }
  for (int i = head[u]; ~i; i = g[i].nxt) {
    int v = g[i].to;
    if (v != pa)
      dp[v][2] = dp[u][1] - used[v] * (dp[v][1] + a[v] - 2ll * g[i].w);
  }
}
void dfsDp3(int u, int pa) {
  suffix[u] = suffix[pa] + dp[u][2];
  for (int i = head[u]; ~i; i = g[i].nxt) {
    int v = g[i].to;
    if (v != pa) {
      dp[v][3] = max(0ll, dp[u][3] + a[u] - 2 * g[i].w + dp[v][2]);
      dfsDp3(v, u);
    }
  }
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = (20), i_end_ = (0); i >= i_end_; --i)
    if (dep[fa[i][u]] >= dep[v]) u = fa[i][u];
  if (u == v) return v;
  for (int i = (20), i_end_ = (0); i >= i_end_; --i)
    if (fa[i][u] != fa[i][v]) u = fa[i][u], v = fa[i][v];
  return fa[0][u];
}
long long calc(int u, int v) {
  int l = LCA(u, v);
  long long rv = valDis[u] + valDis[v] - valDis[l] - valDis[fa[0][l]] -
                 (dis[u] + dis[v] - 2 * dis[l]);
  if (dep[u] > dep[v]) swap(u, v);
  if (l != u) {
    rv += dp[u][1] + dp[v][1] + dp[l][3];
    int param = u, param1 = v;
    for (int i = (20), i_end_ = (0); i >= i_end_; --i) {
      if (dep[fa[i][param]] > dep[l]) param = fa[i][param];
      if (dep[fa[i][param1]] > dep[l]) param1 = fa[i][param1];
    }
    rv += suffix[u] - suffix[param];
    rv += suffix[v] - suffix[param1];
    rv += dp[param1][2];
    if (used[param]) rv -= dp[param][1] + a[param] - 2ll * tov[param];
  } else
    rv += suffix[v] - suffix[u] + dp[u][3] + dp[v][1];
  return rv;
}
void Solve() {
  dfsFa(1, 0);
  dfsDp12(1, 0);
  dfsDp3(1, 0);
  for (int i = (1), i_end_ = (q); i <= i_end_; ++i) {
    int u = read(), v = read();
    printf("%lld\n", calc(u, v));
  }
}
int main() {
  Init();
  Solve();
  return 0;
}