#include <bits/stdc++.h>
const int N = 4e5 + 5;
using namespace std;
struct edge {
  int x, y, z;
} g[N << 1];
long long f[N], p[N], sup[N], sdo[N], fw[N];
int k[N], a[N], fa[N], t[N], si[N], er[N], d[N], fid[N];
int su = 1;
inline int r() {
  char h = getchar();
  int y = 0, q = 1;
  while (h < '0' || h > '9') {
    if (h == '-') q = -1;
    h = getchar();
  }
  while (h >= '0' && h <= '9') y = y * 10 + h - '0', h = getchar();
  return y * q;
}
inline void add(int a, int b, int c) {
  g[++su] = (edge){k[a], b, c};
  k[a] = su;
}
void dfs1(int x, int fa) {
  f[x] = a[x];
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y, z = g[i].z + g[i ^ 1].z;
    if (y == fa) continue;
    dfs1(y, x);
    f[x] += max(f[y] - z, 0ll);
  }
}
void dfs2(int x, int fa, int z) {
  if (fa == 0)
    p[x] = f[x];
  else
    p[x] = f[x] + max(0ll, p[fa] - max(f[x] - z, 0ll) - z);
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y;
    if (y == fa) continue;
    dfs2(y, x, g[i].z + g[i ^ 1].z);
  }
}
void dfs3(int x, int fa) {
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y, z = g[i].z + g[i ^ 1].z;
    if (y == fa) continue;
    sup[y] = sup[x] + f[x] - max(0ll, f[y] - z) - g[i ^ 1].z;
    sdo[y] = sdo[x] + f[x] - max(0ll, f[y] - z) - g[i].z;
    dfs3(y, x);
  }
}
void dfs4(int x, int f) {
  si[x] = 1;
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y;
    if (y == f) continue;
    fa[y] = x;
    d[y] = d[x] + 1;
    fw[y] = g[i].z + g[i ^ 1].z;
    fid[y] = i;
    dfs4(y, x);
    si[x] += si[y];
    if (si[er[x]] < si[y]) er[x] = y;
  }
}
void dfs5(int x, int to) {
  t[x] = to;
  if (!er[x]) return;
  dfs5(er[x], to);
  for (int i = k[x]; i; i = g[i].x)
    if (!t[g[i].y]) dfs5(g[i].y, g[i].y);
}
inline int lca(int x, int y) {
  while (t[x] != t[y]) {
    if (d[t[x]] < d[t[y]]) swap(x, y);
    x = fa[t[x]];
  }
  if (d[x] > d[y]) swap(x, y);
  return x;
}
inline int pre(int x, int l) {
  while (t[x] != t[l] && fa[x] != l) {
    x = t[x];
    if (fa[x] == l) return x;
    x = fa[x];
  }
  if (fa[x] == l) return x;
  return er[l];
}
int main() {
  int n = r(), m = r();
  for (int i = 1; i <= n; i++) a[i] = r();
  for (int a, b, c, d, i = 1; i < n; i++)
    a = r(), b = r(), c = r(), d = c, add(a, b, c), add(b, a, d);
  dfs1(1, 0);
  dfs2(1, 1, 0);
  sup[1] = sdo[1] = 0;
  dfs3(1, 0);
  dfs4(1, 0);
  dfs5(1, 1);
  while (m--) {
    int a = r(), b = r();
    if (a == b)
      printf("%lld\n", p[a]);
    else {
      int l = lca(a, b);
      if (l == 1) {
        printf("%lld\n", sup[a] + sdo[b] + f[a] + f[b] - (f[1]));
      } else {
        int pr = pre(b, l);
        printf("%lld\n",
               sup[a] - sup[l] - max(0ll, f[pr] - fw[pr]) + sdo[b] - sdo[pr] -
                   g[fid[pr]].z +
                   max(0ll, p[fa[l]] - max(f[l] - fw[l], 0ll) - fw[l]) + f[a] +
                   f[b]);
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
const int N = 4e5 + 5;
using namespace std;
struct edge {
  int x, y, z;
} g[N << 1];
long long f[N], p[N], sup[N], sdo[N], fw[N];
int k[N], a[N], fa[N], t[N], si[N], er[N], d[N], fid[N];
int su = 1;
inline int r() {
  char h = getchar();
  int y = 0, q = 1;
  while (h < '0' || h > '9') {
    if (h == '-') q = -1;
    h = getchar();
  }
  while (h >= '0' && h <= '9') y = y * 10 + h - '0', h = getchar();
  return y * q;
}
inline void add(int a, int b, int c) {
  g[++su] = (edge){k[a], b, c};
  k[a] = su;
}
void dfs1(int x, int fa) {
  f[x] = a[x];
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y, z = g[i].z + g[i ^ 1].z;
    if (y == fa) continue;
    dfs1(y, x);
    f[x] += max(f[y] - z, 0ll);
  }
}
void dfs2(int x, int fa, int z) {
  if (fa == 0)
    p[x] = f[x];
  else
    p[x] = f[x] + max(0ll, p[fa] - max(f[x] - z, 0ll) - z);
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y;
    if (y == fa) continue;
    dfs2(y, x, g[i].z + g[i ^ 1].z);
  }
}
void dfs3(int x, int fa) {
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y, z = g[i].z + g[i ^ 1].z;
    if (y == fa) continue;
    sup[y] = sup[x] + f[x] - max(0ll, f[y] - z) - g[i ^ 1].z;
    sdo[y] = sdo[x] + f[x] - max(0ll, f[y] - z) - g[i].z;
    dfs3(y, x);
  }
}
void dfs4(int x, int f) {
  si[x] = 1;
  for (int i = k[x]; i; i = g[i].x) {
    int y = g[i].y;
    if (y == f) continue;
    fa[y] = x;
    d[y] = d[x] + 1;
    fw[y] = g[i].z + g[i ^ 1].z;
    fid[y] = i;
    dfs4(y, x);
    si[x] += si[y];
    if (si[er[x]] < si[y]) er[x] = y;
  }
}
void dfs5(int x, int to) {
  t[x] = to;
  if (!er[x]) return;
  dfs5(er[x], to);
  for (int i = k[x]; i; i = g[i].x)
    if (!t[g[i].y]) dfs5(g[i].y, g[i].y);
}
inline int lca(int x, int y) {
  while (t[x] != t[y]) {
    if (d[t[x]] < d[t[y]]) swap(x, y);
    x = fa[t[x]];
  }
  if (d[x] > d[y]) swap(x, y);
  return x;
}
inline int pre(int x, int l) {
  while (t[x] != t[l] && fa[x] != l) {
    x = t[x];
    if (fa[x] == l) return x;
    x = fa[x];
  }
  if (fa[x] == l) return x;
  return er[l];
}
int main() {
  int n = r(), m = r();
  for (int i = 1; i <= n; i++) a[i] = r();
  for (int a, b, c, d, i = 1; i < n; i++)
    a = r(), b = r(), c = r(), d = c, add(a, b, c), add(b, a, d);
  dfs1(1, 0);
  dfs2(1, 1, 0);
  sup[1] = sdo[1] = 0;
  dfs3(1, 0);
  dfs4(1, 0);
  dfs5(1, 1);
  while (m--) {
    int a = r(), b = r();
    if (a == b)
      printf("%lld\n", p[a]);
    else {
      int l = lca(a, b);
      if (l == 1) {
        printf("%lld\n", sup[a] + sdo[b] + f[a] + f[b] - (f[1]));
      } else {
        int pr = pre(b, l);
        printf("%lld\n",
               sup[a] - sup[l] - max(0ll, f[pr] - fw[pr]) + sdo[b] - sdo[pr] -
                   g[fid[pr]].z +
                   max(0ll, p[fa[l]] - max(f[l] - fw[l], 0ll) - fw[l]) + f[a] +
                   f[b]);
      }
    }
  }
  return 0;
}

```