#include <bits/stdc++.h>
using namespace std;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const int N = 200100;
vector<int> g[N];
long long V[N];
int par[N], imp[N];
int cima[N];
int sz[N];
void dfs_init(int v, int p = -1) {
  sz[v] = 1;
  imp[v] = 0;
  par[v] = 1;
  for (int to : g[v])
    if (to != p) {
      dfs_init(to, v);
      imp[v] += par[to];
      par[v] += imp[to];
      sz[v] += sz[to];
    }
}
long long res = 0;
const int M = 1e9 + 7;
int n;
void dfs(int v, int p = -1) {
  res = (res + 1ll * V[v] * n) % M;
  for (int to : g[v]) {
    int pp = par[to], ii = imp[to];
    int qtd = n - sz[to];
    if (to == p) qtd = sz[v];
    long long cur = mod(V[v] * (qtd), M);
    res = (res + cur * ii) % M;
    res = mod(res - cur * pp, M);
  }
  for (int to : g[v])
    if (to != p) {
      vector<int> antes = {par[v], imp[v], par[to], imp[to]};
      par[v] -= imp[to];
      imp[v] -= par[to];
      par[to] += imp[v];
      imp[to] += par[v];
      dfs(to, v);
      tie(par[v], imp[v], par[to], imp[to]) =
          tie(antes[0], antes[1], antes[2], antes[3]);
    }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> V[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs_init(1);
  dfs(1);
  cout << res << "\n";
}

### 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;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const int N = 200100;
vector<int> g[N];
long long V[N];
int par[N], imp[N];
int cima[N];
int sz[N];
void dfs_init(int v, int p = -1) {
  sz[v] = 1;
  imp[v] = 0;
  par[v] = 1;
  for (int to : g[v])
    if (to != p) {
      dfs_init(to, v);
      imp[v] += par[to];
      par[v] += imp[to];
      sz[v] += sz[to];
    }
}
long long res = 0;
const int M = 1e9 + 7;
int n;
void dfs(int v, int p = -1) {
  res = (res + 1ll * V[v] * n) % M;
  for (int to : g[v]) {
    int pp = par[to], ii = imp[to];
    int qtd = n - sz[to];
    if (to == p) qtd = sz[v];
    long long cur = mod(V[v] * (qtd), M);
    res = (res + cur * ii) % M;
    res = mod(res - cur * pp, M);
  }
  for (int to : g[v])
    if (to != p) {
      vector<int> antes = {par[v], imp[v], par[to], imp[to]};
      par[v] -= imp[to];
      imp[v] -= par[to];
      par[to] += imp[v];
      imp[to] += par[v];
      dfs(to, v);
      tie(par[v], imp[v], par[to], imp[to]) =
          tie(antes[0], antes[1], antes[2], antes[3]);
    }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> V[i];
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs_init(1);
  dfs(1);
  cout << res << "\n";
}

```