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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
template <typename T>
void read(T &a) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  a = x * f;
}
template <typename T>
void print(T x) {
  if (!x) return;
  if (x < 0) putchar('-'), x = -x;
  print(x / 10);
  putchar(x % 10 + '0');
}
template <typename T, typename... Args>
void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
long long n, m, a[300005], head[300005], pnt[300005 << 1], nxt[300005 << 1],
    val[300005 << 1], E = 0;
long long siz[300005], son[300005], dep[300005], fa[300005], top[300005],
    dfn[300005], rnk[300005], cnt = 0;
long long f[300005], g[300005], h[300005], back[300005], pre[300005],
    suf[300005], sum[300005];
void add_edge(long long u, long long v, long long w) {
  val[E] = w;
  pnt[E] = v;
  nxt[E] = head[u];
  head[u] = E++;
}
void dfs1(long long u, long long faa) {
  sum[u] = sum[faa] + a[u];
  fa[u] = faa;
  siz[u] = 1;
  son[u] = -1;
  for (long long i = head[u]; i != -1; i = nxt[i]) {
    long long v = pnt[i];
    if (v == faa) continue;
    back[v] = val[i];
    dep[v] = dep[u] + val[i];
    dfs1(v, u);
    f[u] += h[v];
    siz[u] += siz[v];
    if (son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;
  }
  h[u] = max(0LL, f[u] - 2 * back[u] + a[u]);
}
void dfs2(long long u, long long tp) {
  top[u] = tp;
  dfn[u] = ++cnt;
  rnk[cnt] = u;
  pre[u] = pre[fa[u]] + f[u];
  suf[u] = suf[fa[u]] + h[u];
  g[u] = max(0LL, g[fa[u]] + f[fa[u]] + a[fa[u]] - h[u] - 2 * back[u]);
  if (son[u] == -1) return;
  dfs2(son[u], tp);
  for (long long i = head[u]; i != -1; i = nxt[i]) {
    long long v = pnt[i];
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
long long LCA(long long x, long long y) {
  while (top[x] != top[y]) {
    if (dfn[x] < dfn[y]) swap(x, y);
    x = fa[top[x]];
  }
  if (dfn[x] > dfn[y]) swap(x, y);
  return x;
}
signed main() {
  memset(head, -1, sizeof(head));
  read(n, m);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (long long i = 1; i < n; i++) {
    long long u, v, w;
    read(u, v, w);
    add_edge(u, v, w);
    add_edge(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  while (m--) {
    long long u, v;
    read(u, v);
    long long lca = LCA(u, v);
    long long ans = 0;
    ans += pre[u] + pre[v] - 2 * pre[lca] + f[lca];
    ans -= suf[u] + suf[v] - 2 * suf[lca] - g[lca];
    ans -= dep[u] + dep[v] - 2 * dep[lca];
    ans += sum[u] + sum[v] - sum[lca] - sum[fa[lca]];
    print(ans);
    putchar('\n');
  }
  return 0;
}