#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;
template <class T>
inline void read(T &x) {
  int t;
  bool flag = false;
  while ((t = getchar()) != '-' && (t < '0' || t > '9'))
    ;
  if (t == '-') flag = true, t = getchar();
  x = t - '0';
  while ((t = getchar()) >= '0' && t <= '9') x = x * 10 + t - '0';
  if (flag) x = -x;
}
typedef struct {
  int to, next;
} Edge;
Edge tree[MAXN * 2];
int head[MAXN], cnt;
void add(int from, int to) {
  tree[cnt].to = to;
  tree[cnt].next = head[from];
  head[from] = cnt++;
}
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
long long N, v[MAXN], dp0[MAXN], f0[MAXN], dp2[MAXN], f2[MAXN];
const long long MOD = 1000000007LL;
void dfs0(int root, int par) {
  dp0[root] = v[root], f0[root] = 1;
  for (int i = head[root]; ~i; i = tree[i].next) {
    int to = tree[i].to;
    if (to != par) {
      dfs0(to, root);
      dp0[root] = (f0[to] * v[root] + dp0[root] - dp0[to] + MOD) % MOD;
      dp0[root] = (dp0[root] + MOD) % MOD;
      f0[root] += f0[to];
    }
  }
}
void dfs1(int root, int par) {
  for (int i = head[root]; ~i; i = tree[i].next) {
    int to = tree[i].to;
    if (to != par) {
      dp2[to] = ((N - f0[to]) * v[to] + dp0[to] -
                 (dp2[root] + dp0[to] - f0[to] * v[root] + MOD) + MOD) %
                MOD;
      dp2[to] = (dp2[to] + MOD) % MOD;
      dfs1(to, root);
    }
  }
}
int main() {
  init();
  int a, b;
  read(N);
  for (int i = 1; i <= N; ++i) read(v[i]);
  for (int i = 1; i < N; ++i) {
    read(a), read(b);
    add(a, b);
    add(b, a);
  }
  dfs0(1, -1);
  dp2[1] = dp0[1];
  f2[1] = f0[1];
  dfs1(1, -1);
  long long ans = 0;
  for (int i = 1; i <= N; ++i) ans = (ans + dp2[i]) % MOD;
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 MAXN = 2e5 + 10;
template <class T>
inline void read(T &x) {
  int t;
  bool flag = false;
  while ((t = getchar()) != '-' && (t < '0' || t > '9'))
    ;
  if (t == '-') flag = true, t = getchar();
  x = t - '0';
  while ((t = getchar()) >= '0' && t <= '9') x = x * 10 + t - '0';
  if (flag) x = -x;
}
typedef struct {
  int to, next;
} Edge;
Edge tree[MAXN * 2];
int head[MAXN], cnt;
void add(int from, int to) {
  tree[cnt].to = to;
  tree[cnt].next = head[from];
  head[from] = cnt++;
}
void init() {
  memset(head, -1, sizeof(head));
  cnt = 0;
}
long long N, v[MAXN], dp0[MAXN], f0[MAXN], dp2[MAXN], f2[MAXN];
const long long MOD = 1000000007LL;
void dfs0(int root, int par) {
  dp0[root] = v[root], f0[root] = 1;
  for (int i = head[root]; ~i; i = tree[i].next) {
    int to = tree[i].to;
    if (to != par) {
      dfs0(to, root);
      dp0[root] = (f0[to] * v[root] + dp0[root] - dp0[to] + MOD) % MOD;
      dp0[root] = (dp0[root] + MOD) % MOD;
      f0[root] += f0[to];
    }
  }
}
void dfs1(int root, int par) {
  for (int i = head[root]; ~i; i = tree[i].next) {
    int to = tree[i].to;
    if (to != par) {
      dp2[to] = ((N - f0[to]) * v[to] + dp0[to] -
                 (dp2[root] + dp0[to] - f0[to] * v[root] + MOD) + MOD) %
                MOD;
      dp2[to] = (dp2[to] + MOD) % MOD;
      dfs1(to, root);
    }
  }
}
int main() {
  init();
  int a, b;
  read(N);
  for (int i = 1; i <= N; ++i) read(v[i]);
  for (int i = 1; i < N; ++i) {
    read(a), read(b);
    add(a, b);
    add(b, a);
  }
  dfs0(1, -1);
  dp2[1] = dp0[1];
  f2[1] = f0[1];
  dfs1(1, -1);
  long long ans = 0;
  for (int i = 1; i <= N; ++i) ans = (ans + dp2[i]) % MOD;
  cout << ans << endl;
  return 0;
}

```