#include <bits/stdc++.h>
using namespace std;
void Rd(int &x) {
  static char c;
  x = 0;
  while ((c = getchar()) < 48)
    ;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while ((c = getchar()) > 47);
}
int n;
int Next[200004 << 1], V[200004 << 1], W[200004], Head[200004], tot;
void Add_Edge(int u, int v) { Next[++tot] = Head[u], V[Head[u] = tot] = v; }
int Sz[200004], G, Mn, Nowsz;
bool Mark[200004];
long long Ans, Dis[200004];
void GetG(int x, int f) {
  int y, mx = 0;
  Sz[x] = 1;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]] && y != f) {
      GetG(y, x);
      Sz[x] += Sz[y];
      mx = max(mx, Sz[y]);
    }
  mx = max(mx, Nowsz - Sz[x]);
  if (mx < Mn) Mn = mx, G = x;
}
int m;
struct Node {
  int l;
  long long d;
} D[200004];
long long Calc(int x) {
  long long Tmp[2] = {0, 0}, Cnt[2] = {0, 0}, Res = 0;
  for (int i = (0), i_end_ = (m); i < i_end_; ++i)
    (Tmp[D[i].l & 1] += D[i].d) %= 1000000007, Cnt[D[i].l & 1]++;
  Res -= 2 * Cnt[1] * Tmp[1] + Cnt[1] * Cnt[1] % 1000000007 * W[x];
  Res += 2 * Cnt[0] * Tmp[0] + Cnt[0] * Cnt[0] % 1000000007 * W[x];
  return Res % 1000000007;
}
void Gets(int x, int f, int d) {
  D[m++] = (Node){d, Dis[x]};
  int y;
  Sz[x] = 1;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]] && y != f) {
      Dis[y] = Dis[x] + ((d & 1) ? W[y] : -W[y]);
      Gets(y, x, d + 1);
      Sz[x] += Sz[y];
    }
}
void Work(int x, int sz) {
  Mn = 200004, Nowsz = sz;
  GetG(x, 0);
  Mark[x = G] = 1;
  m = Dis[x] = 0;
  Gets(x, 0, 0);
  Ans += Calc(x);
  if (sz == 1) return;
  int y;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]]) {
      m = 0;
      Gets(y, x, 1);
      Ans -= Calc(x);
      Work(y, Sz[y]);
    }
}
int main() {
  Rd(n);
  for (int i = (1), i_end_ = (n + 1); i < i_end_; ++i) scanf("%d", &W[i]);
  for (int i = (1), i_end_ = (n); i < i_end_; ++i) {
    int u, v, w;
    Rd(u), Rd(v);
    Add_Edge(u, v);
    Add_Edge(v, u);
  }
  Work(1, n);
  printf("%I64d\n", (Ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}

### 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>
using namespace std;
void Rd(int &x) {
  static char c;
  x = 0;
  while ((c = getchar()) < 48)
    ;
  do x = (x << 3) + (x << 1) + (c ^ 48);
  while ((c = getchar()) > 47);
}
int n;
int Next[200004 << 1], V[200004 << 1], W[200004], Head[200004], tot;
void Add_Edge(int u, int v) { Next[++tot] = Head[u], V[Head[u] = tot] = v; }
int Sz[200004], G, Mn, Nowsz;
bool Mark[200004];
long long Ans, Dis[200004];
void GetG(int x, int f) {
  int y, mx = 0;
  Sz[x] = 1;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]] && y != f) {
      GetG(y, x);
      Sz[x] += Sz[y];
      mx = max(mx, Sz[y]);
    }
  mx = max(mx, Nowsz - Sz[x]);
  if (mx < Mn) Mn = mx, G = x;
}
int m;
struct Node {
  int l;
  long long d;
} D[200004];
long long Calc(int x) {
  long long Tmp[2] = {0, 0}, Cnt[2] = {0, 0}, Res = 0;
  for (int i = (0), i_end_ = (m); i < i_end_; ++i)
    (Tmp[D[i].l & 1] += D[i].d) %= 1000000007, Cnt[D[i].l & 1]++;
  Res -= 2 * Cnt[1] * Tmp[1] + Cnt[1] * Cnt[1] % 1000000007 * W[x];
  Res += 2 * Cnt[0] * Tmp[0] + Cnt[0] * Cnt[0] % 1000000007 * W[x];
  return Res % 1000000007;
}
void Gets(int x, int f, int d) {
  D[m++] = (Node){d, Dis[x]};
  int y;
  Sz[x] = 1;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]] && y != f) {
      Dis[y] = Dis[x] + ((d & 1) ? W[y] : -W[y]);
      Gets(y, x, d + 1);
      Sz[x] += Sz[y];
    }
}
void Work(int x, int sz) {
  Mn = 200004, Nowsz = sz;
  GetG(x, 0);
  Mark[x = G] = 1;
  m = Dis[x] = 0;
  Gets(x, 0, 0);
  Ans += Calc(x);
  if (sz == 1) return;
  int y;
  for (int i = Head[x]; i; i = Next[i])
    if (!Mark[y = V[i]]) {
      m = 0;
      Gets(y, x, 1);
      Ans -= Calc(x);
      Work(y, Sz[y]);
    }
}
int main() {
  Rd(n);
  for (int i = (1), i_end_ = (n + 1); i < i_end_; ++i) scanf("%d", &W[i]);
  for (int i = (1), i_end_ = (n); i < i_end_; ++i) {
    int u, v, w;
    Rd(u), Rd(v);
    Add_Edge(u, v);
    Add_Edge(v, u);
  }
  Work(1, n);
  printf("%I64d\n", (Ans % 1000000007 + 1000000007) % 1000000007);
  return 0;
}

```