#include <bits/stdc++.h>
using namespace std;
inline long long int max(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
inline long long int min(long long int a, long long int b, long long int c) {
  return min(min(a, b), c);
}
inline long long int max(long long int a, long long int b) {
  return (a > b) ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return (a < b) ? a : b;
}
inline long long int mul(long long int x, long long int y, long long int mod_) {
  return ((x % mod_) * 1LL * (y % mod_)) % mod_;
}
long long int power(long long int a, long long int n) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long int powm(long long int a, long long int n, long long int mod_) {
  long long int p = 1;
  while (n) {
    if (n % 2) {
      p = mul(p, a, mod_);
    }
    n >>= 1;
    a = mul(a, a, mod_);
  }
  return p % mod_;
}
long long int powi(long long int a, long long int mod_) {
  return powm(a, mod_ - 2, mod_);
}
const int N = 3e5 + 10;
vector<vector<long long int>> graph;
vector<long long int> col, subtree, par, dep, intime, outtime, odd, even, a;
long long int tme = 0, ans = 0;
void dfs(long long int x) {
  col[x] = 1;
  intime[x] = ++tme;
  for (auto y : graph[x]) {
    if (col[y] == 0) {
      par[y] = x;
      dep[y] = dep[x] + 1;
      dfs(y);
      subtree[x] += subtree[y];
      odd[x] += even[y] + 1;
      even[x] += odd[y];
    }
  }
  outtime[x] = ++tme;
}
void ass(long long int n) {
  graph.assign(n, {});
  col.assign(n, 0);
  intime.assign(n, 0);
  outtime.assign(n, 0);
  dep.assign(n, 0);
  par.assign(n, -1);
  subtree.assign(n, 1);
  odd.assign(n, 0);
  even.assign(n, 0);
}
void dfs2(long long int x, long long int codd, long long int ceven) {
  long long int todd = odd[x], teven = even[x];
  if (par[x] != -1) {
    todd += ceven + 1;
    teven += codd;
  }
  long long int val = 0;
  for (auto y : graph[x]) {
    if (y != par[x]) {
      val += (teven - odd[y]) * odd[y];
      val -= (todd - even[y] - 1) * (even[y] + 1);
    } else {
      val += (teven - codd) * codd;
      val -= (todd - ceven - 1) * (ceven + 1);
    }
  }
  val += (2 * teven + 1);
  if (a[x] < 0) a[x] += 1000000007;
  if (val < 0) {
    val += (abs(val) / 1000000007 + 1) * 1000000007;
    val %= 1000000007;
  }
  ans += mul(val, a[x], 1000000007);
  ans %= 1000000007;
  for (auto y : graph[x]) {
    if (y != par[x]) {
      dfs2(y, todd - even[y] - 1, teven - odd[y]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long int mn, mx;
  long long int n, m, t, k, i, j, sum = 0, flag = 0, cnt = 0;
  long long int x = 0, y = 0, z, l, r, q;
  int TC = 1;
  while (TC--) {
    cin >> n;
    a.resize(n);
    for (i = 0; i < n; ++i) cin >> a[i];
    ass(n);
    for (i = 0; i < n - 1; ++i) {
      cin >> x >> y;
      --x;
      --y;
      graph[x].push_back(y);
      graph[y].push_back(x);
    }
    dfs(0);
    dfs2(0, 0, 0);
    cout << ans << '\n';
  };
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
inline long long int max(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
inline long long int min(long long int a, long long int b, long long int c) {
  return min(min(a, b), c);
}
inline long long int max(long long int a, long long int b) {
  return (a > b) ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return (a < b) ? a : b;
}
inline long long int mul(long long int x, long long int y, long long int mod_) {
  return ((x % mod_) * 1LL * (y % mod_)) % mod_;
}
long long int power(long long int a, long long int n) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long int powm(long long int a, long long int n, long long int mod_) {
  long long int p = 1;
  while (n) {
    if (n % 2) {
      p = mul(p, a, mod_);
    }
    n >>= 1;
    a = mul(a, a, mod_);
  }
  return p % mod_;
}
long long int powi(long long int a, long long int mod_) {
  return powm(a, mod_ - 2, mod_);
}
const int N = 3e5 + 10;
vector<vector<long long int>> graph;
vector<long long int> col, subtree, par, dep, intime, outtime, odd, even, a;
long long int tme = 0, ans = 0;
void dfs(long long int x) {
  col[x] = 1;
  intime[x] = ++tme;
  for (auto y : graph[x]) {
    if (col[y] == 0) {
      par[y] = x;
      dep[y] = dep[x] + 1;
      dfs(y);
      subtree[x] += subtree[y];
      odd[x] += even[y] + 1;
      even[x] += odd[y];
    }
  }
  outtime[x] = ++tme;
}
void ass(long long int n) {
  graph.assign(n, {});
  col.assign(n, 0);
  intime.assign(n, 0);
  outtime.assign(n, 0);
  dep.assign(n, 0);
  par.assign(n, -1);
  subtree.assign(n, 1);
  odd.assign(n, 0);
  even.assign(n, 0);
}
void dfs2(long long int x, long long int codd, long long int ceven) {
  long long int todd = odd[x], teven = even[x];
  if (par[x] != -1) {
    todd += ceven + 1;
    teven += codd;
  }
  long long int val = 0;
  for (auto y : graph[x]) {
    if (y != par[x]) {
      val += (teven - odd[y]) * odd[y];
      val -= (todd - even[y] - 1) * (even[y] + 1);
    } else {
      val += (teven - codd) * codd;
      val -= (todd - ceven - 1) * (ceven + 1);
    }
  }
  val += (2 * teven + 1);
  if (a[x] < 0) a[x] += 1000000007;
  if (val < 0) {
    val += (abs(val) / 1000000007 + 1) * 1000000007;
    val %= 1000000007;
  }
  ans += mul(val, a[x], 1000000007);
  ans %= 1000000007;
  for (auto y : graph[x]) {
    if (y != par[x]) {
      dfs2(y, todd - even[y] - 1, teven - odd[y]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long int mn, mx;
  long long int n, m, t, k, i, j, sum = 0, flag = 0, cnt = 0;
  long long int x = 0, y = 0, z, l, r, q;
  int TC = 1;
  while (TC--) {
    cin >> n;
    a.resize(n);
    for (i = 0; i < n; ++i) cin >> a[i];
    ass(n);
    for (i = 0; i < n - 1; ++i) {
      cin >> x >> y;
      --x;
      --y;
      graph[x].push_back(y);
      graph[y].push_back(x);
    }
    dfs(0);
    dfs2(0, 0, 0);
    cout << ans << '\n';
  };
  return 0;
}

```