#include <bits/stdc++.h>
using namespace std;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int a[300105];
vector<pair<int, int> > mp[300105];
long long dp[300105], dpu[300105];
int d[300105], p[300105][20], fw[300105];
void dfs(int x, int pa) {
  d[x] = d[pa] + 1;
  p[x][0] = pa;
  for (auto c : mp[x])
    if (c.first != pa) {
      dfs(c.first, x);
      fw[c.first] = c.second;
      dp[x] += max(dp[c.first] - 2LL * c.second, 0LL);
    }
  dp[x] += a[x];
}
void dfsu(int x, int pa) {
  for (auto c : mp[x])
    if (c.first != pa) {
      int v = c.first, w = c.second;
      long long t = dp[x] + dpu[x] - max(0LL, dp[v] - 2LL * w);
      dpu[v] = max(0LL, t - 2LL * w);
      dfsu(v, x);
    }
}
long long pd[300105], eds[300105];
void dfs2(int x, int pa, long long sum) {
  pd[x] = sum + dp[x];
  eds[x] = eds[pa] + fw[x];
  for (auto c : mp[x])
    if (c.first != pa) {
      int v = c.first, w = c.second;
      dfs2(v, x, sum + dp[x] - max(0LL, dp[v] - 2LL * w));
    }
}
void init() {
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n; i++) {
      p[i][j] = p[p[i][j - 1]][j - 1];
    }
}
int lca(int a, int b) {
  if (d[a] > d[b]) swap(a, b);
  int f = d[b] - d[a];
  for (int i = 0; (1 << i) <= f; i++)
    if ((1 << i) & f) b = p[b][i];
  if (a != b) {
    for (int i = 19; i >= 0; i--)
      if (p[a][i] != p[b][i]) {
        a = p[a][i];
        b = p[b][i];
      }
    a = p[a][0];
  }
  return a;
}
int gbf(int x, int fa) {
  for (int j = 19; ~j; j--) {
    int t = p[x][j];
    if (d[t] <= d[fa]) continue;
    x = t;
  }
  return x;
}
void fmain(int ID) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    mp[u].push_back({v, w});
    mp[v].push_back({u, w});
  }
  dfs(1, 0);
  dfsu(1, 0);
  dfs2(1, 0, 0);
  init();
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int x, y;
    scanf("%d%d", &x, &y);
    long long ans = 0;
    if (x == y)
      ans = dp[x] + dpu[x];
    else {
      if (d[x] < d[y]) swap(x, y);
      int fa = lca(x, y);
      if (fa == y) {
        if (fa == 1) {
          ans = pd[x] - (eds[x] - eds[fa]);
        } else
          ans = pd[x] - (pd[p[fa][0]] - max(0LL, dp[fa] - 2LL * fw[fa])) +
                dpu[fa] - (eds[x] - eds[fa]);
      } else {
        int fx = gbf(x, fa), fy = gbf(y, fa);
        ans = pd[x] - (pd[fa] - max(0LL, dp[fx] - 2LL * fw[fx]));
        ans += pd[y] - (pd[fa] - max(0LL, dp[fy] - 2LL * fw[fy]));
        ans += dpu[fa] + dp[fa] - max(0LL, dp[fx] - 2LL * fw[fx]) -
               max(0LL, dp[fy] - 2LL * fw[fy]);
        ans -= (eds[x] + eds[y] - eds[fa] * 2);
      }
    }
    printf("%lld\n", ans);
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  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;
template <typename T>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 998244353) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
int a[300105];
vector<pair<int, int> > mp[300105];
long long dp[300105], dpu[300105];
int d[300105], p[300105][20], fw[300105];
void dfs(int x, int pa) {
  d[x] = d[pa] + 1;
  p[x][0] = pa;
  for (auto c : mp[x])
    if (c.first != pa) {
      dfs(c.first, x);
      fw[c.first] = c.second;
      dp[x] += max(dp[c.first] - 2LL * c.second, 0LL);
    }
  dp[x] += a[x];
}
void dfsu(int x, int pa) {
  for (auto c : mp[x])
    if (c.first != pa) {
      int v = c.first, w = c.second;
      long long t = dp[x] + dpu[x] - max(0LL, dp[v] - 2LL * w);
      dpu[v] = max(0LL, t - 2LL * w);
      dfsu(v, x);
    }
}
long long pd[300105], eds[300105];
void dfs2(int x, int pa, long long sum) {
  pd[x] = sum + dp[x];
  eds[x] = eds[pa] + fw[x];
  for (auto c : mp[x])
    if (c.first != pa) {
      int v = c.first, w = c.second;
      dfs2(v, x, sum + dp[x] - max(0LL, dp[v] - 2LL * w));
    }
}
void init() {
  for (int j = 1; (1 << j) <= n; j++)
    for (int i = 1; i <= n; i++) {
      p[i][j] = p[p[i][j - 1]][j - 1];
    }
}
int lca(int a, int b) {
  if (d[a] > d[b]) swap(a, b);
  int f = d[b] - d[a];
  for (int i = 0; (1 << i) <= f; i++)
    if ((1 << i) & f) b = p[b][i];
  if (a != b) {
    for (int i = 19; i >= 0; i--)
      if (p[a][i] != p[b][i]) {
        a = p[a][i];
        b = p[b][i];
      }
    a = p[a][0];
  }
  return a;
}
int gbf(int x, int fa) {
  for (int j = 19; ~j; j--) {
    int t = p[x][j];
    if (d[t] <= d[fa]) continue;
    x = t;
  }
  return x;
}
void fmain(int ID) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (int(i) = 1; (i) <= (int)(n - 1); (i)++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    mp[u].push_back({v, w});
    mp[v].push_back({u, w});
  }
  dfs(1, 0);
  dfsu(1, 0);
  dfs2(1, 0, 0);
  init();
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int x, y;
    scanf("%d%d", &x, &y);
    long long ans = 0;
    if (x == y)
      ans = dp[x] + dpu[x];
    else {
      if (d[x] < d[y]) swap(x, y);
      int fa = lca(x, y);
      if (fa == y) {
        if (fa == 1) {
          ans = pd[x] - (eds[x] - eds[fa]);
        } else
          ans = pd[x] - (pd[p[fa][0]] - max(0LL, dp[fa] - 2LL * fw[fa])) +
                dpu[fa] - (eds[x] - eds[fa]);
      } else {
        int fx = gbf(x, fa), fy = gbf(y, fa);
        ans = pd[x] - (pd[fa] - max(0LL, dp[fx] - 2LL * fw[fx]));
        ans += pd[y] - (pd[fa] - max(0LL, dp[fy] - 2LL * fw[fy]));
        ans += dpu[fa] + dp[fa] - max(0LL, dp[fx] - 2LL * fw[fx]) -
               max(0LL, dp[fy] - 2LL * fw[fy]);
        ans -= (eds[x] + eds[y] - eds[fa] * 2);
      }
    }
    printf("%lld\n", ans);
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}

```