#include <bits/stdc++.h>
using namespace std;
long long r;
vector<int> g[200005];
int n;
bool tk[200005];
int fat[200005];
int szt[200005];
long long v[200005];
int calcsz(int x, int f) {
  szt[x] = 1;
  for (auto y : g[x])
    if (y != f && !tk[y]) szt[x] += calcsz(y, x);
  return szt[x];
}
void dfs(int x, int f, int d, long long s, long long* a, int* q) {
  d ^= 1;
  if (d)
    s += v[x];
  else
    s -= v[x];
  a[d] += s;
  q[d]++;
  for (int y : g[x]) {
    if (tk[y] || y == f) continue;
    dfs(y, x, d, s, a, q);
  }
}
void cdfs(int x = 0, int f = -1, int sz = -1) {
  if (sz < 0) sz = calcsz(x, -1);
  for (auto y : g[x])
    if (!tk[y] && szt[y] * 2 >= sz) {
      szt[x] = 0;
      cdfs(y, f, sz);
      return;
    }
  tk[x] = 1;
  fat[x] = f;
  r += v[x];
  long long d[2] = {0, v[x]};
  int q[2] = {0, 1};
  for (int y : g[x]) {
    if (tk[y]) continue;
    long long d2[2] = {0, 0};
    int q2[2] = {0, 0};
    dfs(y, x, 0, 0, d2, q2);
    d2[0] %= 1000000007;
    d2[1] %= 1000000007;
    r += 2 * (q[0] * d2[1] + q2[1] * d[0]);
    r %= 1000000007;
    r += 2 * (q[1] * (1000000007 - d2[0]) + q2[0] * d[1]);
    r %= 1000000007;
    q[0] += q2[1];
    q[1] += q2[0];
    d[0] += (d2[1] - q2[1] * v[x] % 1000000007);
    d[0] %= 1000000007;
    d[1] += ((1000000007 - d2[0]) + q2[0] * v[x] % 1000000007);
    d[1] %= 1000000007;
  }
  for (auto y : g[x])
    if (!tk[y]) cdfs(y, x);
}
void centroid() {
  memset(tk, 0, sizeof(tk));
  cdfs();
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) scanf("%lld", v + i);
  for (int i = 1; i < (n); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  centroid();
  r %= 1000000007;
  r += 1000000007;
  r %= 1000000007;
  printf("%lld\n", r);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
long long r;
vector<int> g[200005];
int n;
bool tk[200005];
int fat[200005];
int szt[200005];
long long v[200005];
int calcsz(int x, int f) {
  szt[x] = 1;
  for (auto y : g[x])
    if (y != f && !tk[y]) szt[x] += calcsz(y, x);
  return szt[x];
}
void dfs(int x, int f, int d, long long s, long long* a, int* q) {
  d ^= 1;
  if (d)
    s += v[x];
  else
    s -= v[x];
  a[d] += s;
  q[d]++;
  for (int y : g[x]) {
    if (tk[y] || y == f) continue;
    dfs(y, x, d, s, a, q);
  }
}
void cdfs(int x = 0, int f = -1, int sz = -1) {
  if (sz < 0) sz = calcsz(x, -1);
  for (auto y : g[x])
    if (!tk[y] && szt[y] * 2 >= sz) {
      szt[x] = 0;
      cdfs(y, f, sz);
      return;
    }
  tk[x] = 1;
  fat[x] = f;
  r += v[x];
  long long d[2] = {0, v[x]};
  int q[2] = {0, 1};
  for (int y : g[x]) {
    if (tk[y]) continue;
    long long d2[2] = {0, 0};
    int q2[2] = {0, 0};
    dfs(y, x, 0, 0, d2, q2);
    d2[0] %= 1000000007;
    d2[1] %= 1000000007;
    r += 2 * (q[0] * d2[1] + q2[1] * d[0]);
    r %= 1000000007;
    r += 2 * (q[1] * (1000000007 - d2[0]) + q2[0] * d[1]);
    r %= 1000000007;
    q[0] += q2[1];
    q[1] += q2[0];
    d[0] += (d2[1] - q2[1] * v[x] % 1000000007);
    d[0] %= 1000000007;
    d[1] += ((1000000007 - d2[0]) + q2[0] * v[x] % 1000000007);
    d[1] %= 1000000007;
  }
  for (auto y : g[x])
    if (!tk[y]) cdfs(y, x);
}
void centroid() {
  memset(tk, 0, sizeof(tk));
  cdfs();
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) scanf("%lld", v + i);
  for (int i = 1; i < (n); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  centroid();
  r %= 1000000007;
  r += 1000000007;
  r %= 1000000007;
  printf("%lld\n", r);
  return 0;
}

```