#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const long long mod = 1e9 + 7;
vector<int> tr[N];
int val[N], used[N], color[N], sz[N], B[N], W[N], p[N];
queue<int> q;
void bfs(int root) {
  int u, v;
  color[root] = 1;
  used[root] = 1;
  p[root] = 0;
  q.push(root);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = 0; i < tr[u].size(); i++) {
      v = tr[u][i];
      if (!used[v]) {
        color[v] = 1 - color[u];
        used[v] = 1;
        p[v] = u;
        q.push(v);
      }
    }
  }
}
void dfs(int x) {
  used[x] = 1;
  for (int i = 0; i < tr[x].size(); i++) {
    if (!used[tr[x][i]]) {
      dfs(tr[x][i]);
      sz[x] += sz[tr[x][i]];
      B[x] += B[tr[x][i]];
      W[x] += W[tr[x][i]];
    }
  }
  if (color[x])
    B[x]++;
  else
    W[x]++;
  sz[x]++;
}
long long getmod(long long x) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
int main() {
  int n, u, v, totB, totW;
  long long ans;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d %d", &u, &v);
    tr[u].push_back(v);
    tr[v].push_back(u);
  }
  memset(used, 0, sizeof(used));
  bfs(1);
  memset(used, 0, sizeof(used));
  memset(sz, 0, sizeof(sz));
  memset(B, 0, sizeof(B));
  memset(W, 0, sizeof(W));
  dfs(1);
  totB = B[1];
  totW = W[1];
  ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += 1LL * val[i] * n % mod;
    ans = getmod(ans);
    if (color[i]) {
      ans += 1LL * val[i] * sz[i] % mod * (totB - B[i] - totW + W[i]) % mod;
      ans = getmod(ans);
      for (int j = 0; j < tr[i].size(); j++) {
        u = tr[i][j];
        if (u != p[i]) {
          ans += 1LL * val[i] * (n - sz[u]) % mod * (B[u] - W[u]) % mod;
          ans = getmod(ans);
        }
      }
    } else {
      ans += 1LL * val[i] * sz[i] % mod * (totW - W[i] - totB + B[i]) % mod;
      ans = getmod(ans);
      for (int j = 0; j < tr[i].size(); j++) {
        u = tr[i][j];
        if (u != p[i]) {
          ans += 1LL * val[i] * (n - sz[u]) % mod * (W[u] - B[u]) % mod;
          ans = getmod(ans);
        }
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### 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;
const int N = 2e5 + 10;
const long long mod = 1e9 + 7;
vector<int> tr[N];
int val[N], used[N], color[N], sz[N], B[N], W[N], p[N];
queue<int> q;
void bfs(int root) {
  int u, v;
  color[root] = 1;
  used[root] = 1;
  p[root] = 0;
  q.push(root);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (int i = 0; i < tr[u].size(); i++) {
      v = tr[u][i];
      if (!used[v]) {
        color[v] = 1 - color[u];
        used[v] = 1;
        p[v] = u;
        q.push(v);
      }
    }
  }
}
void dfs(int x) {
  used[x] = 1;
  for (int i = 0; i < tr[x].size(); i++) {
    if (!used[tr[x][i]]) {
      dfs(tr[x][i]);
      sz[x] += sz[tr[x][i]];
      B[x] += B[tr[x][i]];
      W[x] += W[tr[x][i]];
    }
  }
  if (color[x])
    B[x]++;
  else
    W[x]++;
  sz[x]++;
}
long long getmod(long long x) {
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
int main() {
  int n, u, v, totB, totW;
  long long ans;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d %d", &u, &v);
    tr[u].push_back(v);
    tr[v].push_back(u);
  }
  memset(used, 0, sizeof(used));
  bfs(1);
  memset(used, 0, sizeof(used));
  memset(sz, 0, sizeof(sz));
  memset(B, 0, sizeof(B));
  memset(W, 0, sizeof(W));
  dfs(1);
  totB = B[1];
  totW = W[1];
  ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += 1LL * val[i] * n % mod;
    ans = getmod(ans);
    if (color[i]) {
      ans += 1LL * val[i] * sz[i] % mod * (totB - B[i] - totW + W[i]) % mod;
      ans = getmod(ans);
      for (int j = 0; j < tr[i].size(); j++) {
        u = tr[i][j];
        if (u != p[i]) {
          ans += 1LL * val[i] * (n - sz[u]) % mod * (B[u] - W[u]) % mod;
          ans = getmod(ans);
        }
      }
    } else {
      ans += 1LL * val[i] * sz[i] % mod * (totW - W[i] - totB + B[i]) % mod;
      ans = getmod(ans);
      for (int j = 0; j < tr[i].size(); j++) {
        u = tr[i][j];
        if (u != p[i]) {
          ans += 1LL * val[i] * (n - sz[u]) % mod * (W[u] - B[u]) % mod;
          ans = getmod(ans);
        }
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```