#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
inline void uin(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
inline void uax(T &a, T b) {
  if (b > a) a = b;
}
const long long maxn = 200 * 1000 + 228, mod = 1000 * 1000 * 1000 + 7;
long long n;
vector<long long> g[maxn];
long long a[maxn], sz[maxn], sm[maxn], left228[maxn], right228[maxn], timer = 0;
inline void add(long long &x, long long y) {
  x += y;
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
inline void men(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
inline long long addi(long long a, long long b) {
  add(a, b);
  return a;
}
void zhfs(long long v, long long par = -1) {
  left228[v] = ++timer;
  sz[v] = 1;
  for (long long to : g[v]) {
    if (to != par) {
      zhfs(to, v);
      sz[v] += sz[to];
    }
  }
  right228[v] = left228[v] + sz[v] - 1;
}
void jfs(long long v, long long sum = 0, long long h = 0, long long par = -1) {
  if (h == 0) {
    add(sum, a[v]);
  } else {
    add(sum, -a[v]);
  }
  sm[v] = sum;
  for (long long to : g[v]) {
    if (to != par) {
      jfs(to, sum, h ^ 1, v);
    }
  }
}
long long var = 0;
long long modif = 1;
long long res = 0;
long long mas[maxn];
void add1(long long l, long long r, long long x) {
  long long kek = (r - l + 1) * x;
  if (kek < 0) kek += 200000LL * mod;
  kek %= mod;
  add(var, kek);
}
void minusravno() {
  var = -var;
  if (var < 0) var += mod;
}
long long get1() { return var; }
void dfs(long long v, long long par = -1) {
  if (par != -1) {
    minusravno();
    add1(left228[v], right228[v], a[par]);
    if (left228[v] > 1) add1(1, left228[v] - 1, a[v]);
    if (right228[v] < n) add1(right228[v] + 1, n, a[v]);
  }
  add(res, get1());
  for (long long to : g[v]) {
    if (to != par) {
      dfs(to, v);
    }
  }
  if (par != -1) {
    add1(left228[v], right228[v], -a[par]);
    if (left228[v] > 1) add1(1, left228[v] - 1, -a[v]);
    if (right228[v] < n) add1(right228[v] + 1, n, -a[v]);
    minusravno();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= (long long)n; ++i) cin >> a[i];
  for (long long it = 1; it <= (long long)n - 1; ++it) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  zhfs(1);
  jfs(1);
  var = 0;
  for (long long i = 1; i <= (long long)n; ++i) {
    add1(left228[i], left228[i], sm[i]);
  }
  dfs(1);
  cout << res % mod << "\n";
  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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
inline void uin(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
inline void uax(T &a, T b) {
  if (b > a) a = b;
}
const long long maxn = 200 * 1000 + 228, mod = 1000 * 1000 * 1000 + 7;
long long n;
vector<long long> g[maxn];
long long a[maxn], sz[maxn], sm[maxn], left228[maxn], right228[maxn], timer = 0;
inline void add(long long &x, long long y) {
  x += y;
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
inline void men(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  if (x >= mod) x -= mod;
}
inline long long addi(long long a, long long b) {
  add(a, b);
  return a;
}
void zhfs(long long v, long long par = -1) {
  left228[v] = ++timer;
  sz[v] = 1;
  for (long long to : g[v]) {
    if (to != par) {
      zhfs(to, v);
      sz[v] += sz[to];
    }
  }
  right228[v] = left228[v] + sz[v] - 1;
}
void jfs(long long v, long long sum = 0, long long h = 0, long long par = -1) {
  if (h == 0) {
    add(sum, a[v]);
  } else {
    add(sum, -a[v]);
  }
  sm[v] = sum;
  for (long long to : g[v]) {
    if (to != par) {
      jfs(to, sum, h ^ 1, v);
    }
  }
}
long long var = 0;
long long modif = 1;
long long res = 0;
long long mas[maxn];
void add1(long long l, long long r, long long x) {
  long long kek = (r - l + 1) * x;
  if (kek < 0) kek += 200000LL * mod;
  kek %= mod;
  add(var, kek);
}
void minusravno() {
  var = -var;
  if (var < 0) var += mod;
}
long long get1() { return var; }
void dfs(long long v, long long par = -1) {
  if (par != -1) {
    minusravno();
    add1(left228[v], right228[v], a[par]);
    if (left228[v] > 1) add1(1, left228[v] - 1, a[v]);
    if (right228[v] < n) add1(right228[v] + 1, n, a[v]);
  }
  add(res, get1());
  for (long long to : g[v]) {
    if (to != par) {
      dfs(to, v);
    }
  }
  if (par != -1) {
    add1(left228[v], right228[v], -a[par]);
    if (left228[v] > 1) add1(1, left228[v] - 1, -a[v]);
    if (right228[v] < n) add1(right228[v] + 1, n, -a[v]);
    minusravno();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i <= (long long)n; ++i) cin >> a[i];
  for (long long it = 1; it <= (long long)n - 1; ++it) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  zhfs(1);
  jfs(1);
  var = 0;
  for (long long i = 1; i <= (long long)n; ++i) {
    add1(left228[i], left228[i], sm[i]);
  }
  dfs(1);
  cout << res % mod << "\n";
  return 0;
}

```