#include <bits/stdc++.h>
using namespace std;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int x = 0, f = 1;
  register char ch = nc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = nc();
  }
  while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = nc();
  return x * f;
}
inline void write(register int x) {
  if (!x) putchar('0');
  if (x < 0) x = -x, putchar('-');
  static int sta[20];
  register int tot = 0;
  while (x) sta[tot++] = x % 10, x /= 10;
  while (tot) putchar(sta[--tot] + 48);
}
struct edge {
  int to, nxt, w;
} e[300005 << 1];
int head[300005], cnt;
void add(int u, int v, int w) {
  e[++cnt] = (edge){v, head[u], w};
  head[u] = cnt;
}
int n, q, a[300005], fa[300005][22], dep[300005];
long long f[300005], g[300005];
void dfs1(int x, int fa_) {
  f[x] = a[x];
  fa[x][0] = fa_;
  dep[x] = dep[fa_] + 1;
  for (int i = 1; i <= 20; ++i) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa_) continue;
    dfs1(v, x);
    f[x] += max(0ll, f[v] - 2ll * e[i].w);
  }
}
long long sum1[300005], sum2[300005];
void dfs2(int x) {
  sum1[x] = sum1[fa[x][0]] + f[x];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[x][0]) continue;
    g[v] = max(0ll, f[x] + g[x] - max(0ll, f[v] - 2ll * e[i].w) - 2ll * e[i].w);
    sum2[v] = sum2[x] - e[i].w - max(0ll, f[v] - 2ll * e[i].w);
    dfs2(v);
  }
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 0; i <= 19; ++i)
    if ((1 << i) & (dep[u] - dep[v])) u = fa[u][i];
  for (int i = 19; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return (u == v) ? u : fa[u][0];
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    add(u, v, w), add(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1);
  while (q--) {
    int u = read(), v = read();
    int lca = getlca(u, v);
    long long ans = sum1[u] + sum1[v] - sum1[lca] - sum1[fa[lca][0]] + g[lca] +
                    sum2[u] + sum2[v] - 2 * sum2[lca];
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int read() {
  register int x = 0, f = 1;
  register char ch = nc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = nc();
  }
  while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = nc();
  return x * f;
}
inline void write(register int x) {
  if (!x) putchar('0');
  if (x < 0) x = -x, putchar('-');
  static int sta[20];
  register int tot = 0;
  while (x) sta[tot++] = x % 10, x /= 10;
  while (tot) putchar(sta[--tot] + 48);
}
struct edge {
  int to, nxt, w;
} e[300005 << 1];
int head[300005], cnt;
void add(int u, int v, int w) {
  e[++cnt] = (edge){v, head[u], w};
  head[u] = cnt;
}
int n, q, a[300005], fa[300005][22], dep[300005];
long long f[300005], g[300005];
void dfs1(int x, int fa_) {
  f[x] = a[x];
  fa[x][0] = fa_;
  dep[x] = dep[fa_] + 1;
  for (int i = 1; i <= 20; ++i) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa_) continue;
    dfs1(v, x);
    f[x] += max(0ll, f[v] - 2ll * e[i].w);
  }
}
long long sum1[300005], sum2[300005];
void dfs2(int x) {
  sum1[x] = sum1[fa[x][0]] + f[x];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[x][0]) continue;
    g[v] = max(0ll, f[x] + g[x] - max(0ll, f[v] - 2ll * e[i].w) - 2ll * e[i].w);
    sum2[v] = sum2[x] - e[i].w - max(0ll, f[v] - 2ll * e[i].w);
    dfs2(v);
  }
}
int getlca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 0; i <= 19; ++i)
    if ((1 << i) & (dep[u] - dep[v])) u = fa[u][i];
  for (int i = 19; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return (u == v) ? u : fa[u][0];
}
int main() {
  n = read(), q = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read(), w = read();
    add(u, v, w), add(v, u, w);
  }
  dfs1(1, 0);
  dfs2(1);
  while (q--) {
    int u = read(), v = read();
    int lca = getlca(u, v);
    long long ans = sum1[u] + sum1[v] - sum1[lca] - sum1[fa[lca][0]] + g[lca] +
                    sum2[u] + sum2[v] - 2 * sum2[lca];
    printf("%lld\n", ans);
  }
  return 0;
}

```