#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const pair<T, T> &u) {
  os << u.first << " " << u.second << ' ';
  return os;
}
template <class T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const long long INF = 1e9 + 7;
const long long mod = 998244353;
const long long BIG_INF = 1e18;
const long long N = 1e6 + 7;
const long long T = 1 << 20;
const long double inf = 1e18;
long long n, m, q, SZ;
long long centro;
bool blocked[N];
long long koszt[N];
vector<vector<int> > G(N);
long long dfs(int x, int przed = -1) {
  long long roz = 1;
  bool ok = 1;
  for (auto &u : G[x])
    if (u != przed and !blocked[u]) {
      long long temp = dfs(u, x);
      if (temp + temp > SZ) ok = 0;
      roz += temp;
    }
  if (ok and SZ <= roz * 2) centro = x;
  return roz;
}
long long ans = 0;
long long parz = 0;
long long ile_parz = 0;
long long nparz = koszt[centro];
long long ile_nparz = 1;
void dfs2(int u, int gl, long long suma, int przed) {
  suma += koszt[u];
  if (gl & 1)
    ans += (parz + ile_parz * suma) * 2;
  else
    ans += (nparz + ile_nparz * suma) * 2;
  ans %= INF;
  for (auto &v : G[u])
    if (v != przed and !blocked[v]) dfs2(v, gl + 1, -suma, u);
}
void dfs3(int u, int gl, long long suma, int przed) {
  suma += koszt[u];
  if (gl & 1) {
    parz += suma;
    ile_parz++;
  } else {
    nparz += suma;
    ile_nparz++;
  }
  for (auto &v : G[u])
    if (v != przed and !blocked[v]) dfs3(v, gl + 1, -suma, u);
}
void decomp(int x) {
  SZ = dfs(x), dfs(x);
  int centroid = centro;
  blocked[centroid] = true;
  parz = 0;
  ile_parz = 0;
  nparz = koszt[centroid];
  ile_nparz = 1;
  ans += koszt[centro];
  ans %= INF;
  for (auto &u : G[centroid]) {
    if (!blocked[u]) {
      dfs2(u, 1, 0, centroid);
      dfs3(u, 1, -koszt[centroid], centroid);
    }
  }
  for (auto &u : G[centroid])
    if (!blocked[u]) decomp(u);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> koszt[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  decomp(1);
  cout << (ans + INF) % INF << '\n';
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const pair<T, T> &u) {
  os << u.first << " " << u.second << ' ';
  return os;
}
template <class T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const long long INF = 1e9 + 7;
const long long mod = 998244353;
const long long BIG_INF = 1e18;
const long long N = 1e6 + 7;
const long long T = 1 << 20;
const long double inf = 1e18;
long long n, m, q, SZ;
long long centro;
bool blocked[N];
long long koszt[N];
vector<vector<int> > G(N);
long long dfs(int x, int przed = -1) {
  long long roz = 1;
  bool ok = 1;
  for (auto &u : G[x])
    if (u != przed and !blocked[u]) {
      long long temp = dfs(u, x);
      if (temp + temp > SZ) ok = 0;
      roz += temp;
    }
  if (ok and SZ <= roz * 2) centro = x;
  return roz;
}
long long ans = 0;
long long parz = 0;
long long ile_parz = 0;
long long nparz = koszt[centro];
long long ile_nparz = 1;
void dfs2(int u, int gl, long long suma, int przed) {
  suma += koszt[u];
  if (gl & 1)
    ans += (parz + ile_parz * suma) * 2;
  else
    ans += (nparz + ile_nparz * suma) * 2;
  ans %= INF;
  for (auto &v : G[u])
    if (v != przed and !blocked[v]) dfs2(v, gl + 1, -suma, u);
}
void dfs3(int u, int gl, long long suma, int przed) {
  suma += koszt[u];
  if (gl & 1) {
    parz += suma;
    ile_parz++;
  } else {
    nparz += suma;
    ile_nparz++;
  }
  for (auto &v : G[u])
    if (v != przed and !blocked[v]) dfs3(v, gl + 1, -suma, u);
}
void decomp(int x) {
  SZ = dfs(x), dfs(x);
  int centroid = centro;
  blocked[centroid] = true;
  parz = 0;
  ile_parz = 0;
  nparz = koszt[centroid];
  ile_nparz = 1;
  ans += koszt[centro];
  ans %= INF;
  for (auto &u : G[centroid]) {
    if (!blocked[u]) {
      dfs2(u, 1, 0, centroid);
      dfs3(u, 1, -koszt[centroid], centroid);
    }
  }
  for (auto &u : G[centroid])
    if (!blocked[u]) decomp(u);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> koszt[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  decomp(1);
  cout << (ans + INF) % INF << '\n';
}

```