#include <bits/stdc++.h>
const long long mod = 1e9 + 7;
const int sz = 200010;
std::vector<int> edge[sz];
long long v[sz], c[sz][2];
long long w, b;
long long co[sz], p[sz];
struct ele {
  long long x, y;
};
void dfs(int x, int px) {
  p[x] = px;
  for (int k = 0; k < edge[x].size(); k++) {
    int y = edge[x][k];
    if (px == y) continue;
    co[y] = 1 - co[x];
    if (co[y] == 0)
      w++;
    else
      b++;
    dfs(y, x);
    c[x][0] += c[y][0];
    c[x][1] += c[y][1];
  }
  c[x][co[x]]++;
}
long long ans;
void tru(int n) {
  std::vector<ele> tmp;
  for (int k = 1; k <= n; k++) {
    tmp.clear();
    for (int i = 0; i < edge[k].size(); i++) {
      int y = edge[k][i];
      if (y == p[k]) continue;
      tmp.push_back(ele{c[y][0], c[y][1]});
    }
    tmp.push_back(ele{w - c[k][0], b - c[k][1]});
    long long sx = 0, sy = 0, ti = 0, s2x = 0, s2y = 0;
    for (int i = 0; i < tmp.size(); i++) {
      sx += tmp[i].x;
      sx %= mod;
      sy += tmp[i].y;
      sy %= mod;
      s2x += tmp[i].x * tmp[i].x;
      s2x %= mod;
      s2y += tmp[i].y * tmp[i].y;
      s2y %= mod;
    }
    ti = (sx * sx - s2x) % mod - (sy * sy - s2y) % mod;
    ti %= mod;
    if (co[k] == 1) ti = -ti;
    if (co[k] == 0)
      ti += 2 * sx;
    else
      ti += 2 * sy;
    ti++;
    ti %= mod;
    ans += (ti * v[k]) % mod;
    ans %= mod;
  }
  if (ans < 0) ans += mod;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int k = 1; k <= n; k++) scanf("%lld", &v[k]);
  for (int k = 1; k < n; k++) {
    int x, y;
    scanf("%d%d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  w++;
  dfs(1, 1);
  tru(n);
  printf("%lld", ans);
}

### Prompt

In CPP, your task is to solve 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>
const long long mod = 1e9 + 7;
const int sz = 200010;
std::vector<int> edge[sz];
long long v[sz], c[sz][2];
long long w, b;
long long co[sz], p[sz];
struct ele {
  long long x, y;
};
void dfs(int x, int px) {
  p[x] = px;
  for (int k = 0; k < edge[x].size(); k++) {
    int y = edge[x][k];
    if (px == y) continue;
    co[y] = 1 - co[x];
    if (co[y] == 0)
      w++;
    else
      b++;
    dfs(y, x);
    c[x][0] += c[y][0];
    c[x][1] += c[y][1];
  }
  c[x][co[x]]++;
}
long long ans;
void tru(int n) {
  std::vector<ele> tmp;
  for (int k = 1; k <= n; k++) {
    tmp.clear();
    for (int i = 0; i < edge[k].size(); i++) {
      int y = edge[k][i];
      if (y == p[k]) continue;
      tmp.push_back(ele{c[y][0], c[y][1]});
    }
    tmp.push_back(ele{w - c[k][0], b - c[k][1]});
    long long sx = 0, sy = 0, ti = 0, s2x = 0, s2y = 0;
    for (int i = 0; i < tmp.size(); i++) {
      sx += tmp[i].x;
      sx %= mod;
      sy += tmp[i].y;
      sy %= mod;
      s2x += tmp[i].x * tmp[i].x;
      s2x %= mod;
      s2y += tmp[i].y * tmp[i].y;
      s2y %= mod;
    }
    ti = (sx * sx - s2x) % mod - (sy * sy - s2y) % mod;
    ti %= mod;
    if (co[k] == 1) ti = -ti;
    if (co[k] == 0)
      ti += 2 * sx;
    else
      ti += 2 * sy;
    ti++;
    ti %= mod;
    ans += (ti * v[k]) % mod;
    ans %= mod;
  }
  if (ans < 0) ans += mod;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int k = 1; k <= n; k++) scanf("%lld", &v[k]);
  for (int k = 1; k < n; k++) {
    int x, y;
    scanf("%d%d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  w++;
  dfs(1, 1);
  tru(n);
  printf("%lld", ans);
}

```