#include <bits/stdc++.h>
using namespace std;
const int N = 300005, M = N * 2;
long long read() {
  long long x = 0;
  char ch = getchar();
  while (!('0' <= ch && ch <= '9')) ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  return x;
}
struct Gragh {
  int cnt, y[M], nxt[M], fst[N];
  long long z[M];
  void clear() {
    cnt = 0;
    memset(fst, 0, sizeof fst);
  }
  void add(int a, int b, long long c) {
    y[++cnt] = b, z[cnt] = c, nxt[cnt] = fst[a], fst[a] = cnt;
  }
} g;
int n, q, fa[N][20], depth[N], xx, yy;
long long a[N], dp1[N], f1[N], dp2[N], sum[N], fadis[N], len[N], s[N];
void dfs1(int x, int pre, int d, long long L) {
  fa[x][0] = pre;
  depth[x] = d;
  len[x] = L;
  s[x] = s[pre] + a[x];
  for (int i = 1; i < 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  dp1[x] = 0;
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    long long z = g.z[i];
    if (y != pre) {
      dfs1(y, x, d + 1, L + z);
      fadis[y] = z;
      f1[y] = max(dp1[y] + a[y] - z * 2, 0LL);
      dp1[x] += f1[y];
    }
  }
}
void dfs2(int x, int pre, long long v, long long v2) {
  dp2[x] = v;
  sum[x] = v2;
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    long long z = g.z[i];
    if (y != pre) {
      long long _v = max(v + a[x] + dp1[x] - f1[y] - 2 * z, 0LL);
      long long _v2 = max(v2 + dp1[x] - f1[y], 0LL);
      dfs2(y, x, _v, _v2);
    }
  }
}
int LCA(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (depth[x] - (1 << i) >= depth[y]) x = fa[x][i];
  if (x == y) return x;
  for (int 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() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) a[i] = read();
  g.clear();
  for (int i = 1; i < n; i++) {
    int a = read(), b = read();
    long long c = read();
    g.add(a, b, c);
    g.add(b, a, c);
  }
  dfs1(1, 0, 0, 0);
  dfs2(1, 0, 0, 0);
  while (q--) {
    int x, y, lca;
    scanf("%d%d", &x, &y);
    if (depth[x] > depth[y]) swap(x, y);
    lca = LCA(x, y);
    if (x == lca) {
      if (y == lca) {
        printf("%I64d\n", dp1[x] + dp2[x] + a[x]);
        continue;
      }
      long long ans = s[y] - s[x] + a[x];
      ans -= len[y] - len[x];
      ans += sum[y] - sum[x];
      ans += dp2[x] + dp1[y];
      printf("%I64d\n", ans);
      continue;
    }
    long long ans = s[x] + s[y] - s[lca] * 2 + a[lca];
    ans -= len[x] + len[y] - len[lca] * 2;
    ans += sum[x] + sum[y] - sum[xx] - sum[yy];
    ans += dp1[lca] - f1[xx] - f1[yy];
    ans += dp2[lca] + dp1[x] + dp1[y];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300005, M = N * 2;
long long read() {
  long long x = 0;
  char ch = getchar();
  while (!('0' <= ch && ch <= '9')) ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  return x;
}
struct Gragh {
  int cnt, y[M], nxt[M], fst[N];
  long long z[M];
  void clear() {
    cnt = 0;
    memset(fst, 0, sizeof fst);
  }
  void add(int a, int b, long long c) {
    y[++cnt] = b, z[cnt] = c, nxt[cnt] = fst[a], fst[a] = cnt;
  }
} g;
int n, q, fa[N][20], depth[N], xx, yy;
long long a[N], dp1[N], f1[N], dp2[N], sum[N], fadis[N], len[N], s[N];
void dfs1(int x, int pre, int d, long long L) {
  fa[x][0] = pre;
  depth[x] = d;
  len[x] = L;
  s[x] = s[pre] + a[x];
  for (int i = 1; i < 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  dp1[x] = 0;
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    long long z = g.z[i];
    if (y != pre) {
      dfs1(y, x, d + 1, L + z);
      fadis[y] = z;
      f1[y] = max(dp1[y] + a[y] - z * 2, 0LL);
      dp1[x] += f1[y];
    }
  }
}
void dfs2(int x, int pre, long long v, long long v2) {
  dp2[x] = v;
  sum[x] = v2;
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    long long z = g.z[i];
    if (y != pre) {
      long long _v = max(v + a[x] + dp1[x] - f1[y] - 2 * z, 0LL);
      long long _v2 = max(v2 + dp1[x] - f1[y], 0LL);
      dfs2(y, x, _v, _v2);
    }
  }
}
int LCA(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (depth[x] - (1 << i) >= depth[y]) x = fa[x][i];
  if (x == y) return x;
  for (int 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() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) a[i] = read();
  g.clear();
  for (int i = 1; i < n; i++) {
    int a = read(), b = read();
    long long c = read();
    g.add(a, b, c);
    g.add(b, a, c);
  }
  dfs1(1, 0, 0, 0);
  dfs2(1, 0, 0, 0);
  while (q--) {
    int x, y, lca;
    scanf("%d%d", &x, &y);
    if (depth[x] > depth[y]) swap(x, y);
    lca = LCA(x, y);
    if (x == lca) {
      if (y == lca) {
        printf("%I64d\n", dp1[x] + dp2[x] + a[x]);
        continue;
      }
      long long ans = s[y] - s[x] + a[x];
      ans -= len[y] - len[x];
      ans += sum[y] - sum[x];
      ans += dp2[x] + dp1[y];
      printf("%I64d\n", ans);
      continue;
    }
    long long ans = s[x] + s[y] - s[lca] * 2 + a[lca];
    ans -= len[x] + len[y] - len[lca] * 2;
    ans += sum[x] + sum[y] - sum[xx] - sum[yy];
    ans += dp1[lca] - f1[xx] - f1[yy];
    ans += dp2[lca] + dp1[x] + dp1[y];
    printf("%I64d\n", ans);
  }
  return 0;
}

```