#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long w[MAXN];
class LFS {
 public:
  LFS() {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = 0;
  }
  LFS(int N) {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = N;
  }
  void adde(int from, int to, int w) {
    e[ecnt].to = to;
    e[ecnt].w = w;
    e[ecnt].next = head[from];
    head[from] = ecnt++;
  }
  void addde(int a, int b, int w) {
    adde(a, b, w);
    adde(b, a, w);
  }
  long long solve() {
    memset(dp1, 0, sizeof dp1);
    memset(dp2, 0, sizeof dp2);
    memset(cc, 0, sizeof cc);
    ans = 0;
    dfs1(1, 0);
    dfs2(1, 0);
    return (ans + MOD) % MOD;
  }

 protected:
  struct Edge {
    int to, next, w;
  } e[MAXN * 2];
  int head[MAXN];
  int ecnt;
  int n;
  long long dp1[MAXN][2];
  long long dp2[MAXN][2];
  long long cc[MAXN];
  long long ans;

 private:
  void dfs1(int u, int fa) {
    cc[u] = 0;
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dfs1(v, u);
        dp1[u][0] += dp1[v][1];
        dp1[u][1] += dp1[v][0];
        cc[u] += cc[v];
      }
    }
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        ans += (((cc[u] - cc[v]) * w[u]) % MOD) * (dp1[v][1] - dp1[v][0]) % MOD;
        ans %= MOD;
      }
    }
    ans += (long long)n * w[u] % MOD;
    ans %= MOD;
    ans += (((n - cc[u]) * w[u]) % MOD) * (dp1[u][0] - dp1[u][1]) % MOD;
    ans %= MOD;
    cc[u]++;
    dp1[u][0]++;
  }
  void dfs2(int u, int fa) {
    if (u != 1) {
      ans += (((dp2[u][0] - dp2[u][1]) * w[u]) % MOD) * cc[u] % MOD;
      ans %= MOD;
    }
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dp2[v][0] = dp1[u][1] - dp1[v][0] + dp2[u][1];
        dp2[v][1] = dp1[u][0] - dp1[v][1] + dp2[u][0];
        dfs2(v, u);
      }
    }
  }
};
inline long long read() {
  long long X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    X = (X << 3) + (X << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -X : X;
}
int main() {
  int n = read();
  LFS* tree = new LFS(n);
  for (int i = 1; i <= n; i++) {
    w[i] = read();
  }
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    tree->addde(u, v, 1);
  }
  cout << tree->solve() << endl;
  delete tree;
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int MAXN = 200010;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
long long w[MAXN];
class LFS {
 public:
  LFS() {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = 0;
  }
  LFS(int N) {
    memset(head, -1, sizeof head);
    ecnt = 0;
    n = N;
  }
  void adde(int from, int to, int w) {
    e[ecnt].to = to;
    e[ecnt].w = w;
    e[ecnt].next = head[from];
    head[from] = ecnt++;
  }
  void addde(int a, int b, int w) {
    adde(a, b, w);
    adde(b, a, w);
  }
  long long solve() {
    memset(dp1, 0, sizeof dp1);
    memset(dp2, 0, sizeof dp2);
    memset(cc, 0, sizeof cc);
    ans = 0;
    dfs1(1, 0);
    dfs2(1, 0);
    return (ans + MOD) % MOD;
  }

 protected:
  struct Edge {
    int to, next, w;
  } e[MAXN * 2];
  int head[MAXN];
  int ecnt;
  int n;
  long long dp1[MAXN][2];
  long long dp2[MAXN][2];
  long long cc[MAXN];
  long long ans;

 private:
  void dfs1(int u, int fa) {
    cc[u] = 0;
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dfs1(v, u);
        dp1[u][0] += dp1[v][1];
        dp1[u][1] += dp1[v][0];
        cc[u] += cc[v];
      }
    }
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        ans += (((cc[u] - cc[v]) * w[u]) % MOD) * (dp1[v][1] - dp1[v][0]) % MOD;
        ans %= MOD;
      }
    }
    ans += (long long)n * w[u] % MOD;
    ans %= MOD;
    ans += (((n - cc[u]) * w[u]) % MOD) * (dp1[u][0] - dp1[u][1]) % MOD;
    ans %= MOD;
    cc[u]++;
    dp1[u][0]++;
  }
  void dfs2(int u, int fa) {
    if (u != 1) {
      ans += (((dp2[u][0] - dp2[u][1]) * w[u]) % MOD) * cc[u] % MOD;
      ans %= MOD;
    }
    for (int i = head[u]; i != -1; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        dp2[v][0] = dp1[u][1] - dp1[v][0] + dp2[u][1];
        dp2[v][1] = dp1[u][0] - dp1[v][1] + dp2[u][0];
        dfs2(v, u);
      }
    }
  }
};
inline long long read() {
  long long X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    X = (X << 3) + (X << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -X : X;
}
int main() {
  int n = read();
  LFS* tree = new LFS(n);
  for (int i = 1; i <= n; i++) {
    w[i] = read();
  }
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    tree->addde(u, v, 1);
  }
  cout << tree->solve() << endl;
  delete tree;
  return 0;
}

```