#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 65, mod = 1e9 + 7;
vector<int> g[N];
int w[N];
int col[N], pr[N];
int sz[N][2];
long long ans = 0;
void dfs(int v, int par = -1, int c = 0) {
  pr[v] = par;
  ++sz[v][c];
  col[v] = c;
  for (int u : g[v])
    if (u != par) {
      dfs(u, v, c ^ 1);
      sz[v][0] += sz[u][0];
      sz[v][1] += sz[u][1];
    }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", w + i);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  ans = 0;
  dfs(0);
  for (int v = 0; v < n; ++v) {
    vector<pair<int, int> > vs;
    for (int u : g[v])
      if (u != pr[v]) vs.emplace_back(sz[u][0], sz[u][1]);
    if (pr[v] != -1) vs.emplace_back(sz[0][0] - sz[v][0], sz[0][1] - sz[v][1]);
    long long cnt = n;
    if (col[v]) {
      for (pair<int, int> e : vs) {
        cnt = (cnt + 1LL * e.second * (n - e.first - e.second)) % mod;
        cnt = (cnt - 1LL * e.first * (n - e.first - e.second)) % mod;
      }
    } else {
      for (pair<int, int> e : vs) {
        cnt = (cnt - 1LL * e.second * (n - e.first - e.second)) % mod;
        cnt = (cnt + 1LL * e.first * (n - e.first - e.second)) % mod;
      }
    }
    cnt %= mod;
    if (cnt < 0) cnt += mod;
    ans = (ans + cnt * w[v]) % mod;
  }
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 2e5 + 65, mod = 1e9 + 7;
vector<int> g[N];
int w[N];
int col[N], pr[N];
int sz[N][2];
long long ans = 0;
void dfs(int v, int par = -1, int c = 0) {
  pr[v] = par;
  ++sz[v][c];
  col[v] = c;
  for (int u : g[v])
    if (u != par) {
      dfs(u, v, c ^ 1);
      sz[v][0] += sz[u][0];
      sz[v][1] += sz[u][1];
    }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", w + i);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  ans = 0;
  dfs(0);
  for (int v = 0; v < n; ++v) {
    vector<pair<int, int> > vs;
    for (int u : g[v])
      if (u != pr[v]) vs.emplace_back(sz[u][0], sz[u][1]);
    if (pr[v] != -1) vs.emplace_back(sz[0][0] - sz[v][0], sz[0][1] - sz[v][1]);
    long long cnt = n;
    if (col[v]) {
      for (pair<int, int> e : vs) {
        cnt = (cnt + 1LL * e.second * (n - e.first - e.second)) % mod;
        cnt = (cnt - 1LL * e.first * (n - e.first - e.second)) % mod;
      }
    } else {
      for (pair<int, int> e : vs) {
        cnt = (cnt - 1LL * e.second * (n - e.first - e.second)) % mod;
        cnt = (cnt + 1LL * e.first * (n - e.first - e.second)) % mod;
      }
    }
    cnt %= mod;
    if (cnt < 0) cnt += mod;
    ans = (ans + cnt * w[v]) % mod;
  }
  ans %= mod;
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}

```