#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline int read(T& a) {
  a = 0;
  char c = getchar(), f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    if (c == -1) return -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    a = (a << 1) + (a << 3) + (c ^ 48), c = getchar();
  a *= f;
  return 1;
}
template <typename T, typename... Args>
inline int read(T& x, Args&... args) {
  return read(x) + read(args...);
}
template <typename T>
inline int write(T x) {
  if (x < 0) return putchar('-') & write(-x);
  if (x / 10) return write(x / 10) & putchar(x % 10 | 48);
  return putchar(x | 48);
}
template <typename T>
inline int write(T x, char c) {
  return write(x) & putchar(c);
}
long long n;
struct edge {
  long long v, nxt;
} e[200005 << 1];
long long f[200005], tot;
inline void addEdge(long long u, long long v) {
  tot++, e[tot].v = v, e[tot].nxt = f[u], f[u] = tot;
}
inline void init() {
  tot = 0;
  memset(e, 0, sizeof e);
  memset(f, 0, sizeof f);
}
inline void dfs(long long x, long long fa);
inline void bfs();
long long a[200005];
long long s[200005][2], sup[200005][2], siz[200005], fa[200005];
long long ans;
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (long long i = 1, u, v; i < n; i++)
    read(u, v), addEdge(u, v), addEdge(v, u);
  dfs(1, -1);
  bfs();
  write((ans + 1000000007) % 1000000007, '\n');
}
inline void dfs(long long x, long long fat) {
  siz[x] = 1;
  s[x][0] = 1;
  for (long long i = f[x]; i; i = e[i].nxt) {
    long long v = e[i].v;
    if (v == fat) continue;
    fa[v] = x;
    dfs(v, x);
    siz[x] += siz[v];
    s[x][0] += s[v][1], s[x][1] += s[v][0];
  }
  ans += 1ll * (s[x][0] - s[x][1]) * a[x] % 1000000007 * (n - siz[x] + 1) %
         1000000007,
      ans %= 1000000007;
  for (long long i = f[x]; i; i = e[i].nxt) {
    long long v = e[i].v;
    if (v == fat) continue;
    ans += 1ll * (s[v][1] - s[v][0]) % 1000000007 * a[x] % 1000000007 *
           (siz[x] - siz[v] - 1) % 1000000007,
        ans %= 1000000007;
  }
  ans += 1ll * a[x] * (siz[x] - 1) % 1000000007, ans %= 1000000007;
}
inline void bfs() {
  queue<long long> q;
  q.push(1);
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    if (x != 1) {
      sup[x][0] = sup[fa[x]][1] + (s[fa[x]][1] - s[x][0]);
      sup[x][1] = sup[fa[x]][0] + (s[fa[x]][0] - s[x][1]);
    }
    ans +=
        1ll * (sup[x][0] - sup[x][1]) * a[x] % 1000000007 * siz[x] % 1000000007,
        ans %= 1000000007;
    for (long long i = f[x]; i; i = e[i].nxt)
      if (e[i].v != fa[x]) q.push(e[i].v);
  }
}

### 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;
template <typename T>
inline int read(T& a) {
  a = 0;
  char c = getchar(), f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    if (c == -1) return -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    a = (a << 1) + (a << 3) + (c ^ 48), c = getchar();
  a *= f;
  return 1;
}
template <typename T, typename... Args>
inline int read(T& x, Args&... args) {
  return read(x) + read(args...);
}
template <typename T>
inline int write(T x) {
  if (x < 0) return putchar('-') & write(-x);
  if (x / 10) return write(x / 10) & putchar(x % 10 | 48);
  return putchar(x | 48);
}
template <typename T>
inline int write(T x, char c) {
  return write(x) & putchar(c);
}
long long n;
struct edge {
  long long v, nxt;
} e[200005 << 1];
long long f[200005], tot;
inline void addEdge(long long u, long long v) {
  tot++, e[tot].v = v, e[tot].nxt = f[u], f[u] = tot;
}
inline void init() {
  tot = 0;
  memset(e, 0, sizeof e);
  memset(f, 0, sizeof f);
}
inline void dfs(long long x, long long fa);
inline void bfs();
long long a[200005];
long long s[200005][2], sup[200005][2], siz[200005], fa[200005];
long long ans;
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++) read(a[i]);
  for (long long i = 1, u, v; i < n; i++)
    read(u, v), addEdge(u, v), addEdge(v, u);
  dfs(1, -1);
  bfs();
  write((ans + 1000000007) % 1000000007, '\n');
}
inline void dfs(long long x, long long fat) {
  siz[x] = 1;
  s[x][0] = 1;
  for (long long i = f[x]; i; i = e[i].nxt) {
    long long v = e[i].v;
    if (v == fat) continue;
    fa[v] = x;
    dfs(v, x);
    siz[x] += siz[v];
    s[x][0] += s[v][1], s[x][1] += s[v][0];
  }
  ans += 1ll * (s[x][0] - s[x][1]) * a[x] % 1000000007 * (n - siz[x] + 1) %
         1000000007,
      ans %= 1000000007;
  for (long long i = f[x]; i; i = e[i].nxt) {
    long long v = e[i].v;
    if (v == fat) continue;
    ans += 1ll * (s[v][1] - s[v][0]) % 1000000007 * a[x] % 1000000007 *
           (siz[x] - siz[v] - 1) % 1000000007,
        ans %= 1000000007;
  }
  ans += 1ll * a[x] * (siz[x] - 1) % 1000000007, ans %= 1000000007;
}
inline void bfs() {
  queue<long long> q;
  q.push(1);
  while (!q.empty()) {
    long long x = q.front();
    q.pop();
    if (x != 1) {
      sup[x][0] = sup[fa[x]][1] + (s[fa[x]][1] - s[x][0]);
      sup[x][1] = sup[fa[x]][0] + (s[fa[x]][0] - s[x][1]);
    }
    ans +=
        1ll * (sup[x][0] - sup[x][1]) * a[x] % 1000000007 * siz[x] % 1000000007,
        ans %= 1000000007;
    for (long long i = f[x]; i; i = e[i].nxt)
      if (e[i].v != fa[x]) q.push(e[i].v);
  }
}

```