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;
void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = v * 10 + c - '0';
  while (isdigit(c = getchar())) v = v * 10 + c - '0';
  x = v * f;
}
void readc(char &x) {
  char c;
  while ((c = getchar()) == ' ')
    ;
  x = c;
}
void writes(string s) { puts(s.c_str()); }
void writeln() { writes(""); }
void writei(int x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
void writell(long long x) {
  if (x < 0) {
    putchar('-');
    x = abs(x);
  }
  if (!x) putchar('0');
  char a[25];
  int top = 0;
  while (x) {
    a[++top] = (x % 10) + '0';
    x /= 10;
  }
  while (top) {
    putchar(a[top]);
    top--;
  }
}
inline long long inc(int &x) { return ++x; }
inline long long inc(long long &x) { return ++x; }
inline long long inc(int &x, long long y) { return x += y; }
inline long long inc(long long &x, long long y) { return x += y; }
inline double inc(double &x, double y) { return x += y; }
inline long long dec(int &x) { return --x; }
inline long long dec(long long &x) { return --x; }
inline long long dec(int &x, long long y) { return x -= y; }
inline long long dec(long long &x, long long y) { return x -= y; }
inline double dec(double &x, double y) { return x -= y; }
inline long long mul(int &x) { return x = ((long long)x) * x; }
inline long long mul(long long &x) { return x = x * x; }
inline long long mul(int &x, long long y) { return x *= y; }
inline long long mul(long long &x, long long y) { return x *= y; }
inline double mul(double &x, double y) { return x *= y; }
inline long long divi(const int &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(const long long &x) {
  long long ans, l, r, mid;
  ans = 0;
  l = 0;
  r = 0x3fffffff;
  while (l < r) {
    mid = (l + r) / 2;
    if (mid * mid <= x) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid;
  }
  return ans;
}
inline long long divi(int &x, long long y) { return x /= y; }
inline long long divi(long long &x, long long y) { return x /= y; }
inline double divi(double &x, double y) { return x /= y; }
inline long long mod(int &x, long long y) { return x %= y; }
inline long long mod(long long &x, long long y) { return x %= y; }
long long n, m, i, j, dp1[300005], dp2[300005], f1[300005], f2[300005],
    dep[300005], dis[300005], s[300005], x, y, fa[300005][20], z, v[300005], xx,
    yy;
vector<pair<long long, long long> > e[300005];
void dfs1(int x, int p) {
  fa[x][0] = p;
  s[x] += v[x];
  for (__typeof((e[x]).begin()) it = (e[x]).begin(); it != (e[x]).end(); it++) {
    if (it->first == p) continue;
    dis[it->first] = dis[x] + it->second;
    s[it->first] = s[x];
    dep[it->first] = dep[x] + 1;
    dfs1(it->first, x);
    f1[it->first] = max(0ll, dp1[it->first] + v[it->first] - 2 * it->second);
    dp1[x] += f1[it->first];
  }
}
void dfs2(int x, int p) {
  for (__typeof((e[x]).begin()) it = (e[x]).begin(); it != (e[x]).end(); it++) {
    if (it->first == p) continue;
    dp2[it->first] =
        max(0ll, dp2[x] + v[x] + dp1[x] - f1[it->first] - 2 * it->second);
    f2[it->first] = max(0ll, f2[x] + dp1[x] - f1[it->first]);
    dfs2(it->first, x);
  }
}
int lca(int x, int y, long long &xx, long long &yy) {
  if (dep[x] > dep[y]) swap(x, y);
  int t = dep[y] - dep[x], i;
  for (i = 0; i <= 19; i++)
    if ((t >> i) & 1) {
      y = fa[y][i];
    }
  if (x == y) return x;
  for (i = 19; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) {
      x = fa[x][i];
      y = fa[y][i];
    }
  xx = x;
  yy = y;
  return fa[x][0];
}
int main() {
  read(n);
  read(m);
  if ((1) <= ((n)))
    for (((i)) = (1); ((i)) <= ((n)); ((i))++) {
      read(v[i]);
    }
  if ((1) <= ((n - 1)))
    for (((i)) = (1); ((i)) <= ((n - 1)); ((i))++) {
      scanf("%lld%lld%lld", &x, &y, &z);
      e[x].push_back(make_pair(y, z));
      e[y].push_back(make_pair(x, z));
    }
  dfs1(1, 0);
  dfs2(1, 0);
  if ((1) <= ((19)))
    for (((j)) = (1); ((j)) <= ((19)); ((j))++)
      if ((1) <= ((n)))
        for (((i)) = (1); ((i)) <= ((n)); ((i))++)
          fa[i][j] = fa[fa[i][j - 1]][j - 1];
  while (m--) {
    scanf("%lld%lld", &x, &y);
    if (x == y) {
      printf("%lld\n", v[x] + dp1[x] + dp2[x]);
      continue;
    }
    if (dep[x] > dep[y]) swap(x, y);
    z = lca(x, y, xx, yy);
    if (z == x) {
      long long ans = s[y] - s[x] + v[x];
      ans -= dis[y] - dis[x];
      ans += f2[y] - f2[x];
      ans += dp1[y] + dp2[x];
      printf("%lld\n", ans);
      continue;
    }
    long long ans = s[y] + s[x] + v[z] - 2 * s[z];
    ans -= dis[y] + dis[x] - 2 * dis[z];
    ans += f2[y] + f2[x] - f2[yy] - f2[xx];
    ans += dp1[z] - f1[xx] - f1[yy];
    ans += dp2[z] + dp1[x] + dp1[y];
    printf("%lld\n", ans);
  }
  return 0;
}