#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <class T>
void mxi(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void mni(T& a, const T& b) {
  a = min(a, b);
}
long double EPS = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<long long> > g;
long long n, dp[200200][2], odp[200200][2], ans, v[200200];
void dfs(long long nn, long long pp) {
  dp[nn][1]++;
  for (auto x : g[nn]) {
    if (x != pp) {
      dfs(x, nn);
      for (long long i = 0; i < (2); i++) dp[nn][i] += dp[x][i ^ 1];
    }
  }
}
void dfs2(long long nn, long long pp) {
  if (pp) {
    for (long long i = 0; i < (2); i++)
      odp[nn][i] += (odp[pp][i ^ 1] + dp[pp][i ^ 1] - dp[nn][i]);
  }
  for (auto x : g[nn])
    if (x != pp) dfs2(x, nn);
}
void dfs3(long long nn, long long pp) {
  long long odd = 1, even = 0;
  for (auto x : g[nn]) {
    if (x != pp) {
      odd += dp[x][0];
      even += dp[x][1];
    } else {
      odd += odp[pp][0];
      even += odp[pp][1];
    }
  }
  for (auto x : g[nn]) {
    if (x != pp) {
      long long change = (dp[x][0] * (n - dp[x][0] - dp[x][1])) % 1000000007;
      change = (change * v[nn]) % 1000000007;
      ans = (ans + change) % 1000000007;
      change = (dp[x][1] * (n - dp[x][0] - dp[x][1])) % 1000000007;
      change = (-change * v[nn]) % 1000000007;
      change += 1000000007;
      change %= 1000000007;
      ans = (ans + change) % 1000000007;
      dfs3(x, nn);
    } else {
      long long change =
          (odp[nn][1] * (n - odp[nn][0] - odp[nn][1])) % 1000000007;
      change = (change * v[nn]) % 1000000007;
      ans = (ans + change) % 1000000007;
      change = (odp[nn][0] * (n - odp[nn][0] - odp[nn][1])) % 1000000007;
      change = (-change * v[nn]) % 1000000007;
      change += 1000000007;
      change %= 1000000007;
      ans = (ans + change) % 1000000007;
    }
  }
  ans = (ans + (v[nn] * n) % 1000000007) % 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
}
void solve() {
  cin >> n;
  g.resize(n + 1);
  for (long long i = 0; i < (n); i++) cin >> v[i + 1];
  for (long long i = 0; i < (n - 1); i++) {
    long long u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, tab;
  t = 1;
  tab = t;
  while (t--) {
    solve();
  }
  cerr << ((double)clock() / CLOCKS_PER_SEC);
}

### Prompt

In CPP, your task is to solve 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <class T>
void mxi(T& a, const T& b) {
  a = max(a, b);
}
template <class T>
void mni(T& a, const T& b) {
  a = min(a, b);
}
long double EPS = 1e-9;
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<long long> > g;
long long n, dp[200200][2], odp[200200][2], ans, v[200200];
void dfs(long long nn, long long pp) {
  dp[nn][1]++;
  for (auto x : g[nn]) {
    if (x != pp) {
      dfs(x, nn);
      for (long long i = 0; i < (2); i++) dp[nn][i] += dp[x][i ^ 1];
    }
  }
}
void dfs2(long long nn, long long pp) {
  if (pp) {
    for (long long i = 0; i < (2); i++)
      odp[nn][i] += (odp[pp][i ^ 1] + dp[pp][i ^ 1] - dp[nn][i]);
  }
  for (auto x : g[nn])
    if (x != pp) dfs2(x, nn);
}
void dfs3(long long nn, long long pp) {
  long long odd = 1, even = 0;
  for (auto x : g[nn]) {
    if (x != pp) {
      odd += dp[x][0];
      even += dp[x][1];
    } else {
      odd += odp[pp][0];
      even += odp[pp][1];
    }
  }
  for (auto x : g[nn]) {
    if (x != pp) {
      long long change = (dp[x][0] * (n - dp[x][0] - dp[x][1])) % 1000000007;
      change = (change * v[nn]) % 1000000007;
      ans = (ans + change) % 1000000007;
      change = (dp[x][1] * (n - dp[x][0] - dp[x][1])) % 1000000007;
      change = (-change * v[nn]) % 1000000007;
      change += 1000000007;
      change %= 1000000007;
      ans = (ans + change) % 1000000007;
      dfs3(x, nn);
    } else {
      long long change =
          (odp[nn][1] * (n - odp[nn][0] - odp[nn][1])) % 1000000007;
      change = (change * v[nn]) % 1000000007;
      ans = (ans + change) % 1000000007;
      change = (odp[nn][0] * (n - odp[nn][0] - odp[nn][1])) % 1000000007;
      change = (-change * v[nn]) % 1000000007;
      change += 1000000007;
      change %= 1000000007;
      ans = (ans + change) % 1000000007;
    }
  }
  ans = (ans + (v[nn] * n) % 1000000007) % 1000000007;
  ans += 1000000007;
  ans %= 1000000007;
}
void solve() {
  cin >> n;
  g.resize(n + 1);
  for (long long i = 0; i < (n); i++) cin >> v[i + 1];
  for (long long i = 0; i < (n - 1); i++) {
    long long u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  dfs(1, 0);
  dfs2(1, 0);
  dfs3(1, 0);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, tab;
  t = 1;
  tab = t;
  while (t--) {
    solve();
  }
  cerr << ((double)clock() / CLOCKS_PER_SEC);
}

```