#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
long long val[maxn], a[maxn], dp[maxn];
vector<pair<int, long long> > adj[maxn];
long long mp[maxn];
long long g[maxn];
inline void dfs0(int u, int f) {
  long long ans = 0;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    dfs0(v, u);
    ans += max(0ll, dp[v] - w * 2);
  }
  dp[u] = a[u] + ans;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    mp[v] = dp[u] - max(0ll, dp[v] - w * 2);
  }
}
int anc[maxn][21], dep[maxn];
long long sum[maxn][21];
inline void dfs1(int u, int fa) {
  anc[u][0] = fa;
  dep[u] = dep[fa] + 1;
  sum[u][0] = mp[u] - val[fa];
  for (int i = 1; i <= 20; i++) {
    anc[u][i] = anc[anc[u][i - 1]][i - 1];
    sum[u][i] = sum[anc[u][i - 1]][i - 1] + sum[u][i - 1];
  }
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == fa) continue;
    val[v] = w;
    dfs1(v, u);
  }
}
inline void dfs2(int u, int f) {
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    g[v] = max(0ll, mp[v] + max(0ll, g[u] - val[u] * 2));
    dfs2(v, u);
  }
}
int n, m;
inline void init() {
  dfs0(1, 0);
  dfs1(1, 0);
  dfs2(1, 0);
}
inline int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 20; i >= 0; i--) {
    if (dep[anc[u][i]] >= dep[v]) u = anc[u][i];
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; i--) {
    if (anc[u][i] != anc[v][i]) {
      u = anc[u][i];
      v = anc[v][i];
    }
  }
  return anc[u][0];
}
inline long long work(int u, int v) {
  if (u == v) {
    return dp[u] + max(0ll, g[u] - val[u] * 2);
  }
  int z = lca(u, v);
  if (z == u || z == v) {
    if (z == u) swap(u, v);
    long long ans = 0;
    int cur = u;
    for (int i = 20; i >= 0; i--) {
      if (dep[anc[cur][i]] >= dep[z] + 1) {
        ans += sum[cur][i];
        cur = anc[cur][i];
      }
    }
    ans += dp[u] - val[u];
    ans += mp[cur] + max(0ll, g[z] - val[z] * 2);
    return ans;
  } else {
    long long ans = dp[u] - val[u] + dp[v] - val[v];
    for (int i = 20; i >= 0; i--) {
      if (dep[anc[u][i]] >= dep[z] + 1) {
        ans += sum[u][i];
        u = anc[u][i];
      }
      if (dep[anc[v][i]] >= dep[z] + 1) {
        ans += sum[v][i];
        v = anc[v][i];
      }
    }
    ans += dp[z] - max(0ll, dp[u] - val[u] * 2) - max(0ll, dp[v] - val[v] * 2) +
           max(0ll, g[z] - val[z] * 2);
    return ans;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    long long w;
    scanf("%d%d%lld", &u, &v, &w);
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
  }
  init();
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%lld\n", work(u, v));
  }
}

### Prompt

Create a solution in cpp for 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 maxn = 3e5 + 5;
long long val[maxn], a[maxn], dp[maxn];
vector<pair<int, long long> > adj[maxn];
long long mp[maxn];
long long g[maxn];
inline void dfs0(int u, int f) {
  long long ans = 0;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    dfs0(v, u);
    ans += max(0ll, dp[v] - w * 2);
  }
  dp[u] = a[u] + ans;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    mp[v] = dp[u] - max(0ll, dp[v] - w * 2);
  }
}
int anc[maxn][21], dep[maxn];
long long sum[maxn][21];
inline void dfs1(int u, int fa) {
  anc[u][0] = fa;
  dep[u] = dep[fa] + 1;
  sum[u][0] = mp[u] - val[fa];
  for (int i = 1; i <= 20; i++) {
    anc[u][i] = anc[anc[u][i - 1]][i - 1];
    sum[u][i] = sum[anc[u][i - 1]][i - 1] + sum[u][i - 1];
  }
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == fa) continue;
    val[v] = w;
    dfs1(v, u);
  }
}
inline void dfs2(int u, int f) {
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i].first;
    long long w = adj[u][i].second;
    if (v == f) continue;
    g[v] = max(0ll, mp[v] + max(0ll, g[u] - val[u] * 2));
    dfs2(v, u);
  }
}
int n, m;
inline void init() {
  dfs0(1, 0);
  dfs1(1, 0);
  dfs2(1, 0);
}
inline int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 20; i >= 0; i--) {
    if (dep[anc[u][i]] >= dep[v]) u = anc[u][i];
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; i--) {
    if (anc[u][i] != anc[v][i]) {
      u = anc[u][i];
      v = anc[v][i];
    }
  }
  return anc[u][0];
}
inline long long work(int u, int v) {
  if (u == v) {
    return dp[u] + max(0ll, g[u] - val[u] * 2);
  }
  int z = lca(u, v);
  if (z == u || z == v) {
    if (z == u) swap(u, v);
    long long ans = 0;
    int cur = u;
    for (int i = 20; i >= 0; i--) {
      if (dep[anc[cur][i]] >= dep[z] + 1) {
        ans += sum[cur][i];
        cur = anc[cur][i];
      }
    }
    ans += dp[u] - val[u];
    ans += mp[cur] + max(0ll, g[z] - val[z] * 2);
    return ans;
  } else {
    long long ans = dp[u] - val[u] + dp[v] - val[v];
    for (int i = 20; i >= 0; i--) {
      if (dep[anc[u][i]] >= dep[z] + 1) {
        ans += sum[u][i];
        u = anc[u][i];
      }
      if (dep[anc[v][i]] >= dep[z] + 1) {
        ans += sum[v][i];
        v = anc[v][i];
      }
    }
    ans += dp[z] - max(0ll, dp[u] - val[u] * 2) - max(0ll, dp[v] - val[v] * 2) +
           max(0ll, g[z] - val[z] * 2);
    return ans;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    long long w;
    scanf("%d%d%lld", &u, &v, &w);
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
  }
  init();
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    printf("%lld\n", work(u, v));
  }
}

```