#include <bits/stdc++.h>
using namespace std;
const int N = 300005, inf = 0x3f3f3f3f;
int read() {
  int f = 1, g = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) g = g * 10 + ch - '0';
  return f * g;
}
int n, m, fa[N][22], a[N], depth[N];
long long fw[N][22], f[N], g[N][22], h[N];
int size, fir[N];
struct edge {
  int u, v, w, nex;
} e[N * 2];
void add(int u, int v, int w) {
  e[++size] = (edge){u, v, w, fir[u]};
  fir[u] = size;
}
long long mi(long long s, int x) { return s - max(f[x] - fw[x][0] * 2, 0ll); }
void build(int x, int fath) {
  fa[x][0] = fath;
  f[x] = a[x];
  depth[x] = depth[fath] + 1;
  for (int i = 1; i <= 20; i++) {
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
    fw[x][i] = fw[fa[x][i - 1]][i - 1] + fw[x][i - 1];
  }
  for (int i = fir[x]; i; i = e[i].nex)
    if (e[i].v != fath) {
      fw[e[i].v][0] = e[i].w;
      build(e[i].v, x);
      if (f[e[i].v] - e[i].w - e[i].w > 0) f[x] += f[e[i].v] - e[i].w - e[i].w;
    }
}
void build2(int x, int fath) {
  h[x] = f[x];
  if (mi(h[fa[x][0]], x) - fw[x][0] * 2 > 0)
    h[x] += mi(h[fa[x][0]], x) - fw[x][0] * 2;
  for (int i = fir[x]; i; i = e[i].nex)
    if (e[i].v != fath) build2(e[i].v, x);
}
int lca(int x, int y) {
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= depth[y]) x = fa[x][i];
  for (int i = 20; i >= 0; i--)
    if (depth[fa[y][i]] >= depth[x]) y = fa[y][i];
  if (x == y) return x;
  for (int i = 20; i >= 0; i--)
    if (fa[y][i] != fa[x][i]) {
      y = fa[y][i];
      x = fa[x][i];
    }
  return fa[x][0];
}
int upto(int x, int dep) {
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= dep) x = fa[x][i];
  return x;
}
long long getw(int u, int v) {
  long long t = 0;
  for (int i = 20; i >= 0; i--)
    if (depth[fa[u][i]] >= depth[v]) {
      t += fw[u][i];
      u = fa[u][i];
    }
  return t;
}
long long calc(int x, int dep) {
  long long t = 0;
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= dep) {
      t += g[x][i];
      x = fa[x][i];
    }
  return t;
}
int main() {
  n = read();
  m = 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);
  }
  build(1, 0);
  for (int i = 1; i <= n; i++) g[i][0] = mi(f[fa[i][0]], i);
  build2(1, 0);
  for (int k = 1; k <= 20; k++)
    for (int i = 0; i <= n; i++) g[i][k] = g[fa[i][k - 1]][k - 1] + g[i][k - 1];
  while (m--) {
    int u = read(), v = read(), p = lca(u, v);
    if (u == v) {
      printf("%I64d\n", h[u]);
      continue;
    }
    if (depth[u] > depth[v]) swap(u, v);
    long long ans = mi(h[p], upto(v, depth[p] + 1)) + f[v] - getw(v, p) +
                    calc(v, depth[p] + 1);
    if (u != p)
      ans = mi(ans, upto(u, depth[p] + 1)) + f[u] - getw(u, p) +
            calc(u, depth[p] + 1);
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 300005, inf = 0x3f3f3f3f;
int read() {
  int f = 1, g = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) g = g * 10 + ch - '0';
  return f * g;
}
int n, m, fa[N][22], a[N], depth[N];
long long fw[N][22], f[N], g[N][22], h[N];
int size, fir[N];
struct edge {
  int u, v, w, nex;
} e[N * 2];
void add(int u, int v, int w) {
  e[++size] = (edge){u, v, w, fir[u]};
  fir[u] = size;
}
long long mi(long long s, int x) { return s - max(f[x] - fw[x][0] * 2, 0ll); }
void build(int x, int fath) {
  fa[x][0] = fath;
  f[x] = a[x];
  depth[x] = depth[fath] + 1;
  for (int i = 1; i <= 20; i++) {
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
    fw[x][i] = fw[fa[x][i - 1]][i - 1] + fw[x][i - 1];
  }
  for (int i = fir[x]; i; i = e[i].nex)
    if (e[i].v != fath) {
      fw[e[i].v][0] = e[i].w;
      build(e[i].v, x);
      if (f[e[i].v] - e[i].w - e[i].w > 0) f[x] += f[e[i].v] - e[i].w - e[i].w;
    }
}
void build2(int x, int fath) {
  h[x] = f[x];
  if (mi(h[fa[x][0]], x) - fw[x][0] * 2 > 0)
    h[x] += mi(h[fa[x][0]], x) - fw[x][0] * 2;
  for (int i = fir[x]; i; i = e[i].nex)
    if (e[i].v != fath) build2(e[i].v, x);
}
int lca(int x, int y) {
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= depth[y]) x = fa[x][i];
  for (int i = 20; i >= 0; i--)
    if (depth[fa[y][i]] >= depth[x]) y = fa[y][i];
  if (x == y) return x;
  for (int i = 20; i >= 0; i--)
    if (fa[y][i] != fa[x][i]) {
      y = fa[y][i];
      x = fa[x][i];
    }
  return fa[x][0];
}
int upto(int x, int dep) {
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= dep) x = fa[x][i];
  return x;
}
long long getw(int u, int v) {
  long long t = 0;
  for (int i = 20; i >= 0; i--)
    if (depth[fa[u][i]] >= depth[v]) {
      t += fw[u][i];
      u = fa[u][i];
    }
  return t;
}
long long calc(int x, int dep) {
  long long t = 0;
  for (int i = 20; i >= 0; i--)
    if (depth[fa[x][i]] >= dep) {
      t += g[x][i];
      x = fa[x][i];
    }
  return t;
}
int main() {
  n = read();
  m = 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);
  }
  build(1, 0);
  for (int i = 1; i <= n; i++) g[i][0] = mi(f[fa[i][0]], i);
  build2(1, 0);
  for (int k = 1; k <= 20; k++)
    for (int i = 0; i <= n; i++) g[i][k] = g[fa[i][k - 1]][k - 1] + g[i][k - 1];
  while (m--) {
    int u = read(), v = read(), p = lca(u, v);
    if (u == v) {
      printf("%I64d\n", h[u]);
      continue;
    }
    if (depth[u] > depth[v]) swap(u, v);
    long long ans = mi(h[p], upto(v, depth[p] + 1)) + f[v] - getw(v, p) +
                    calc(v, depth[p] + 1);
    if (u != p)
      ans = mi(ans, upto(u, depth[p] + 1)) + f[u] - getw(u, p) +
            calc(u, depth[p] + 1);
    printf("%I64d\n", ans);
  }
  return 0;
}

```