#include <bits/stdc++.h>
namespace FastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
template <class T, class... U>
bool read(T& h, U&... t) {
  return read(h) && read(t...);
}
};  // namespace FastIO
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
long long dp_up[N][2], dp_down[N][2];
long long ans, n, w[N];
vector<long long> e[N];
void dfs1(long long u, long long fa) {
  dp_down[u][0] = 1;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dfs1(v, u);
    dp_down[u][0] += dp_down[v][1];
    dp_down[u][1] += dp_down[v][0];
  }
}
void dfs2(long long u, long long fa) {
  ans = (ans + w[u] * n % mod + mod) % mod;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dp_up[v][0] = dp_up[u][1] + dp_down[u][1] - dp_down[v][0];
    dp_up[v][1] = dp_up[u][0] + dp_down[u][0] - dp_down[v][1];
    ans = (ans +
           w[v] * dp_up[v][0] % mod * (dp_down[v][1] + dp_down[v][0]) % mod +
           mod) %
          mod;
    ans = (ans -
           w[v] * dp_up[v][1] % mod * (dp_down[v][1] + dp_down[v][0]) % mod +
           mod) %
          mod;
    ans =
        (ans + w[u] * dp_down[v][1] % mod * (dp_up[v][1] + dp_up[v][0]) % mod +
         mod) %
        mod;
    ans =
        (ans - w[u] * dp_down[v][0] % mod * (dp_up[v][1] + dp_up[v][0]) % mod +
         mod) %
        mod;
    dfs2(v, u);
  }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> w[i];
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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>
namespace FastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T& x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
template <class T, class... U>
bool read(T& h, U&... t) {
  return read(h) && read(t...);
}
};  // namespace FastIO
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
long long dp_up[N][2], dp_down[N][2];
long long ans, n, w[N];
vector<long long> e[N];
void dfs1(long long u, long long fa) {
  dp_down[u][0] = 1;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dfs1(v, u);
    dp_down[u][0] += dp_down[v][1];
    dp_down[u][1] += dp_down[v][0];
  }
}
void dfs2(long long u, long long fa) {
  ans = (ans + w[u] * n % mod + mod) % mod;
  for (auto v : e[u]) {
    if (v == fa) continue;
    dp_up[v][0] = dp_up[u][1] + dp_down[u][1] - dp_down[v][0];
    dp_up[v][1] = dp_up[u][0] + dp_down[u][0] - dp_down[v][1];
    ans = (ans +
           w[v] * dp_up[v][0] % mod * (dp_down[v][1] + dp_down[v][0]) % mod +
           mod) %
          mod;
    ans = (ans -
           w[v] * dp_up[v][1] % mod * (dp_down[v][1] + dp_down[v][0]) % mod +
           mod) %
          mod;
    ans =
        (ans + w[u] * dp_down[v][1] % mod * (dp_up[v][1] + dp_up[v][0]) % mod +
         mod) %
        mod;
    ans =
        (ans - w[u] * dp_down[v][0] % mod * (dp_up[v][1] + dp_up[v][0]) % mod +
         mod) %
        mod;
    dfs2(v, u);
  }
}
signed main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> w[i];
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  cout << ans << endl;
  return 0;
}

```