#include <bits/stdc++.h>
using namespace std;
int n;
int lab[200005];
vector<vector<int> > a(200005);
int child[200005];
bool dead[200005];
const int mod = 1e9 + 7;
void pdfs(int u, int p) {
  child[u] = 1;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    pdfs(v, u);
    child[u] += child[v];
  }
}
int findcen(int u, int p, const int total) {
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    if (child[v] > total / 2) return findcen(v, u, total);
  }
  return u;
}
long long curSum[2], preSum[2];
long long ans = 0;
void dfs(int u, int p, int c, const int key, const int root, const int out) {
  curSum[c] = (curSum[c] + lab[u]) % mod;
  if (key == 0)
    ans = (ans +
           (curSum[c] - curSum[c ^ 1] + (1 - c * 2) * lab[root]) * 1LL * out +
           preSum[c] - preSum[c ^ 1]) %
          mod;
  else
    preSum[c] = (preSum[c] + 1LL * (curSum[c] - curSum[c ^ 1]) * key) % mod;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    dfs(v, u, c ^ 1, key, root, out);
  }
  curSum[c] = (curSum[c] - lab[u]) % mod;
}
void fdfs(int u, int p, int c, int root) {
  curSum[c] = (curSum[c] + lab[u]) % mod;
  ans = (ans + curSum[0] - curSum[1] + lab[root] + curSum[c] - curSum[c ^ 1] +
         (1 - c * 2) * lab[root]) %
        mod;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    fdfs(v, u, c ^ 1, root);
  }
  curSum[c] = (curSum[c] - lab[u]) % mod;
}
void cal(int u, int phase) {
  int buff = 0;
  if (phase == 0) {
    for (int i = 0; i < a[u].size(); i++) {
      int v = a[u][i];
      if (dead[v]) continue;
      dfs(v, u, 1, 0, u, buff);
      dfs(v, u, 1, 1, u, buff);
      buff += child[v];
    }
  } else {
    for (int i = a[u].size() - 1; i >= 0; i--) {
      int v = a[u][i];
      if (dead[v]) continue;
      dfs(v, u, 1, 0, u, buff);
      dfs(v, u, 1, 1, u, buff);
      buff += child[v];
    }
  }
  memset(curSum, 0, sizeof(curSum));
  memset(preSum, 0, sizeof(preSum));
}
void solve(int u) {
  pdfs(u, u);
  u = findcen(u, u, child[u]);
  pdfs(u, u);
  ans = (ans + lab[u]) % mod;
  dead[u] = true;
  for (auto v : a[u]) {
    if (dead[v]) continue;
    fdfs(v, u, 1, u);
  }
  cal(u, 0);
  cal(u, 1);
  for (auto v : a[u]) {
    if (dead[v]) continue;
    solve(v);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> lab[i];
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  solve(1);
  ans %= mod;
  while (ans < 0) ans += mod;
  cout << ans << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n;
int lab[200005];
vector<vector<int> > a(200005);
int child[200005];
bool dead[200005];
const int mod = 1e9 + 7;
void pdfs(int u, int p) {
  child[u] = 1;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    pdfs(v, u);
    child[u] += child[v];
  }
}
int findcen(int u, int p, const int total) {
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    if (child[v] > total / 2) return findcen(v, u, total);
  }
  return u;
}
long long curSum[2], preSum[2];
long long ans = 0;
void dfs(int u, int p, int c, const int key, const int root, const int out) {
  curSum[c] = (curSum[c] + lab[u]) % mod;
  if (key == 0)
    ans = (ans +
           (curSum[c] - curSum[c ^ 1] + (1 - c * 2) * lab[root]) * 1LL * out +
           preSum[c] - preSum[c ^ 1]) %
          mod;
  else
    preSum[c] = (preSum[c] + 1LL * (curSum[c] - curSum[c ^ 1]) * key) % mod;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    dfs(v, u, c ^ 1, key, root, out);
  }
  curSum[c] = (curSum[c] - lab[u]) % mod;
}
void fdfs(int u, int p, int c, int root) {
  curSum[c] = (curSum[c] + lab[u]) % mod;
  ans = (ans + curSum[0] - curSum[1] + lab[root] + curSum[c] - curSum[c ^ 1] +
         (1 - c * 2) * lab[root]) %
        mod;
  for (auto v : a[u]) {
    if (v == p || dead[v]) continue;
    fdfs(v, u, c ^ 1, root);
  }
  curSum[c] = (curSum[c] - lab[u]) % mod;
}
void cal(int u, int phase) {
  int buff = 0;
  if (phase == 0) {
    for (int i = 0; i < a[u].size(); i++) {
      int v = a[u][i];
      if (dead[v]) continue;
      dfs(v, u, 1, 0, u, buff);
      dfs(v, u, 1, 1, u, buff);
      buff += child[v];
    }
  } else {
    for (int i = a[u].size() - 1; i >= 0; i--) {
      int v = a[u][i];
      if (dead[v]) continue;
      dfs(v, u, 1, 0, u, buff);
      dfs(v, u, 1, 1, u, buff);
      buff += child[v];
    }
  }
  memset(curSum, 0, sizeof(curSum));
  memset(preSum, 0, sizeof(preSum));
}
void solve(int u) {
  pdfs(u, u);
  u = findcen(u, u, child[u]);
  pdfs(u, u);
  ans = (ans + lab[u]) % mod;
  dead[u] = true;
  for (auto v : a[u]) {
    if (dead[v]) continue;
    fdfs(v, u, 1, u);
  }
  cal(u, 0);
  cal(u, 1);
  for (auto v : a[u]) {
    if (dead[v]) continue;
    solve(v);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> lab[i];
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  solve(1);
  ans %= mod;
  while (ans < 0) ans += mod;
  cout << ans << '\n';
}

```