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 MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 3e5 + 5;
const int logn = 20;
int n, q;
int a[maxn];
vector<pair<int, int> > adj[maxn];
int ww[maxn];
long long f[maxn];
long long g[maxn];
long long h[maxn];
int lev[maxn];
int par[logn][maxn];
long long ch[maxn];
long long sa[maxn];
long long sb[maxn];
long long dfs(int u, int p) {
  for (pair<int, int> e : adj[u]) {
    int v = e.first;
    int w = e.second;
    if (v != p) {
      f[u] += max(0LL, dfs(v, u) + a[v] - w - w);
      ww[v] = w;
    }
  }
  return f[u];
}
void upd_dfs(int u, int p, long long val) {
  int pw = 0;
  long long t = 0;
  for (pair<int, int> e : adj[u]) {
    int v = e.first;
    int w = e.second;
    if (v == p) {
      pw = w;
      t += max(0LL, val + a[p] - pw - pw);
    } else {
      t += max(0LL, f[v] + a[v] - w - w);
    }
  }
  if (pw) {
    g[u] = max(0LL, val + a[p] - pw - pw);
  }
  if (pw) {
    h[u] = f[p] - max(0LL, f[u] + a[u] - pw - pw);
  }
  for (pair<int, int> e : adj[u]) {
    int v = e.first;
    int w = e.second;
    if (v != p) {
      upd_dfs(v, u, t - max(0LL, f[v] + a[v] - w - w));
    }
  }
}
void build_dfs(int u, int p) {
  for (int i = (1); i < (logn); ++i) par[i][u] = par[i - 1][par[i - 1][u]];
  for (pair<int, int> e : adj[u]) {
    int v = e.first;
    int w = e.second;
    if (v != p) {
      lev[v] = lev[u] + 1;
      par[0][v] = u;
      sa[v] = sa[u] + a[v];
      sb[v] = sb[u] + w;
      ch[v] = ch[u] + h[v];
      build_dfs(v, u);
    }
  }
}
int lca(int u, int v) {
  if (lev[u] < lev[v]) swap(u, v);
  for (int i = (logn)-1; i >= (0); --i) {
    if (lev[par[i][u]] >= lev[v]) {
      u = par[i][u];
    }
  }
  if (u == v) return u;
  for (int i = (logn)-1; i >= (0); --i) {
    if (par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  }
  return par[0][u];
}
int jump(int u, int p) {
  for (int i = (logn)-1; i >= (0); --i) {
    if (lev[par[i][u]] > lev[p]) {
      u = par[i][u];
    }
  }
  return u;
}
long long query(int u, int v) {
  if (u == v) {
    return a[u] + f[u] + g[u];
  }
  int p = lca(u, v);
  long long res = sa[u] + sa[v] - sa[p];
  if (p) res -= sa[par[0][p]];
  res -= sb[u] + sb[v] - sb[p] - sb[p];
  int uu, vv;
  if (u != p) {
    uu = jump(u, p);
    res += ch[u];
    if (uu) res -= ch[par[0][uu]];
  }
  if (v != p) {
    vv = jump(v, p);
    res += ch[v];
    if (vv) res -= ch[par[0][vv]];
  }
  if (u != p && v != p) {
    res -= f[p];
    res += f[u] + f[v];
  }
  if (u == p) {
    res += f[v];
  }
  if (v == p) {
    res += f[u];
  }
  res += g[p];
  return res;
}
void chemthan() {
  cin >> n >> q;
  for (int i = (0); i < (n); ++i) cin >> a[i];
  for (int i = (0); i < (n - 1); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
  }
  dfs(0, -1), upd_dfs(0, -1, 0);
  sa[0] = a[0], ch[0] = h[0], build_dfs(0, -1);
  while (q--) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    cout << query(u, v) << "\n";
  }
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}