#include <bits/stdc++.h>
using namespace std;
using li = long long;
using ld = long double;
void solve(bool);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  solve(true);
  return 0;
}
vector<vector<li>> e;
vector<li> value;
vector<array<li, 2>> sum, cnt;
li answer;
const li MOD = 1e9 + 7;
void add(li& to, li x) {
  assert(x >= 0 && x < MOD);
  assert(to >= 0 && to < MOD);
  to += x;
  if (to >= MOD) to -= MOD;
}
void dfs(li v, li p) {
  add(sum[v][1], value[v]);
  add(cnt[v][1], 1);
  for (li u : e[v]) {
    if (u == p) continue;
    dfs(u, v);
    for (li t = 0; t < 2; ++t) {
      li s = (t == 0 ? 1 : MOD - 1);
      s = 1;
      add(answer, (sum[v][t] * cnt[u][1 - t] +
                   s * ((sum[u][1 - t] * cnt[v][t]) % MOD)) %
                      MOD);
    }
    for (li t = 0; t < 2; ++t) {
      add(sum[v][t], (sum[u][1 - t] +
                      cnt[u][1 - t] * (t == 0 ? MOD - value[v] : value[v])) %
                         MOD);
      add(cnt[v][t], cnt[u][1 - t]);
    }
  }
}
li stress = 0;
li sdfs(li v, li p, li cur_s, li cur_sign) {
  cur_s += value[v] * cur_sign;
  cur_sign *= -1;
  li add = cur_s;
  for (li u : e[v]) {
    if (u == p) continue;
    add += sdfs(u, v, cur_s, cur_sign);
  }
  return add;
}
void solve(bool __attribute__((unused)) read) {
  li n;
  if (read) {
    cin >> n;
  } else {
    n = rand() % 10 + 1;
    cout << "TEST:\n";
    cout << n << endl;
  }
  value.assign(n, 0);
  li sum_vals = 0;
  for (li i = 0; i < n; ++i) {
    if (read) {
      cin >> value[i];
    } else {
      value[i] = rand() % 5 + 1;
      cout << value[i] << " ";
    }
    value[i] %= MOD;
    if (value[i] < 0) value[i] += MOD;
    add(sum_vals, value[i]);
  }
  if (!read) {
    cout << endl;
  }
  e.clear();
  e.resize(n);
  for (li i = 0; i + 1 < n; ++i) {
    li u, v;
    if (read)
      cin >> u >> v;
    else {
      u = rand() % (i + 1);
      v = i + 1;
      ++u;
      ++v;
      cout << u << " " << v << endl;
    }
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  array<li, 2> zero;
  zero.fill(0);
  sum.assign(n, zero);
  cnt.assign(n, zero);
  answer = 0;
  dfs(0, -1);
  add(answer, answer);
  add(answer, sum_vals % MOD);
  cout << answer << endl;
}

### Prompt

Generate a cpp solution to the following problem:

Given a tree with n nodes numbered from 1 to n. Each node i has an associated value V_i.

If the simple path from u_1 to u_m consists of m nodes namely u_1 → u_2 → u_3 → ... u_{m-1} → u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = ∑_{i=1}^{m} (-1)^{i+1} ⋅ V_{u_{i}}. A path can also have 0 edges, i.e. u_{1}=u_{m}.

Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo 10^{9}+7. 

Input

The first line contains an integer n (2 ≤ n ≤ 2⋅10^{5} ) — the number of vertices in the tree.

The second line contains n space-separated integers V_1, V_2, …, V_n (-10^9≤ V_i ≤ 10^9) — values of the nodes.

The next n-1 lines each contain two space-separated integers u and v (1≤ u, v≤ n, u ≠ v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.

Output

Print the total sum of alternating functions of all unique simple paths modulo 10^{9}+7. 

Examples

Input

4
-4 1 5 -2
1 2
1 3
1 4


Output

40


Input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8


Output

4

Note

Consider the first example.

A simple path from node 1 to node 2: 1 → 2 has alternating function equal to A(1,2) = 1 ⋅ (-4)+(-1) ⋅ 1 = -5.

A simple path from node 1 to node 3: 1 → 3 has alternating function equal to A(1,3) = 1 ⋅ (-4)+(-1) ⋅ 5 = -9.

A simple path from node 2 to node 4: 2 → 1 → 4 has alternating function A(2,4) = 1 ⋅ (1)+(-1) ⋅ (-4)+1 ⋅ (-2) = 3.

A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 ⋅ (-4) = -4.

Similarly, A(2, 1) = 5, A(3, 1) = 9, A(4, 2) = 3, A(1, 4) = -2, A(4, 1) = 2, A(2, 2) = 1, A(3, 3) = 5, A(4, 4) = -2, A(3, 4) = 7, A(4, 3) = 7, A(2, 3) = 10, A(3, 2) = 10. So the answer is (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40.

Similarly A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 which sums upto 40. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using li = long long;
using ld = long double;
void solve(bool);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(20);
  solve(true);
  return 0;
}
vector<vector<li>> e;
vector<li> value;
vector<array<li, 2>> sum, cnt;
li answer;
const li MOD = 1e9 + 7;
void add(li& to, li x) {
  assert(x >= 0 && x < MOD);
  assert(to >= 0 && to < MOD);
  to += x;
  if (to >= MOD) to -= MOD;
}
void dfs(li v, li p) {
  add(sum[v][1], value[v]);
  add(cnt[v][1], 1);
  for (li u : e[v]) {
    if (u == p) continue;
    dfs(u, v);
    for (li t = 0; t < 2; ++t) {
      li s = (t == 0 ? 1 : MOD - 1);
      s = 1;
      add(answer, (sum[v][t] * cnt[u][1 - t] +
                   s * ((sum[u][1 - t] * cnt[v][t]) % MOD)) %
                      MOD);
    }
    for (li t = 0; t < 2; ++t) {
      add(sum[v][t], (sum[u][1 - t] +
                      cnt[u][1 - t] * (t == 0 ? MOD - value[v] : value[v])) %
                         MOD);
      add(cnt[v][t], cnt[u][1 - t]);
    }
  }
}
li stress = 0;
li sdfs(li v, li p, li cur_s, li cur_sign) {
  cur_s += value[v] * cur_sign;
  cur_sign *= -1;
  li add = cur_s;
  for (li u : e[v]) {
    if (u == p) continue;
    add += sdfs(u, v, cur_s, cur_sign);
  }
  return add;
}
void solve(bool __attribute__((unused)) read) {
  li n;
  if (read) {
    cin >> n;
  } else {
    n = rand() % 10 + 1;
    cout << "TEST:\n";
    cout << n << endl;
  }
  value.assign(n, 0);
  li sum_vals = 0;
  for (li i = 0; i < n; ++i) {
    if (read) {
      cin >> value[i];
    } else {
      value[i] = rand() % 5 + 1;
      cout << value[i] << " ";
    }
    value[i] %= MOD;
    if (value[i] < 0) value[i] += MOD;
    add(sum_vals, value[i]);
  }
  if (!read) {
    cout << endl;
  }
  e.clear();
  e.resize(n);
  for (li i = 0; i + 1 < n; ++i) {
    li u, v;
    if (read)
      cin >> u >> v;
    else {
      u = rand() % (i + 1);
      v = i + 1;
      ++u;
      ++v;
      cout << u << " " << v << endl;
    }
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  array<li, 2> zero;
  zero.fill(0);
  sum.assign(n, zero);
  cnt.assign(n, zero);
  answer = 0;
  dfs(0, -1);
  add(answer, answer);
  add(answer, sum_vals % MOD);
  cout << answer << endl;
}

```