#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
long long n, m, k, sz[N], a[N], f[N], g[N], ans, cnt, head[N];
struct node {
  long long to, next;
} edge[N << 1];
inline void add(long long u, long long v) {
  edge[++cnt] = {v, head[u]};
  head[u] = cnt;
}
inline void dfs(long long u, long long fa) {
  f[u] = sz[u] = 1;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    f[u] += g[v];
    g[u] += f[v];
  }
  ans = (ans + (f[u] - g[u]) * (n - sz[u] + 1) % mod * a[u] % mod) % mod;
  ans = (ans + (sz[u] - 1) * a[u] % mod) % mod;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    ans = (ans + (g[v] - f[v]) * (sz[u] - sz[v] - 1) % mod * a[u] % mod) % mod;
  }
}
long long uf[N], ug[N];
inline void work(long long u, long long fa) {
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    uf[v] = ug[u] + g[u] - f[v];
    ug[v] = uf[u] + f[u] - g[v];
    ans = (ans + (uf[v] - ug[v]) * sz[v] % mod * a[v] % mod) % mod;
    work(v, u);
  }
}
signed main() {
  n = read();
  long long u, v;
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i < n; i++) {
    u = read(), v = read();
    add(u, v), add(v, u);
  }
  dfs(1, 0);
  work(1, 0);
  cout << (ans + mod) % mod;
  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;
template <typename T>
void chmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <typename T>
void chmax(T &x, const T &y) {
  if (x < y) x = y;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const long long N = 2e5 + 10;
const long long mod = 1e9 + 7;
long long n, m, k, sz[N], a[N], f[N], g[N], ans, cnt, head[N];
struct node {
  long long to, next;
} edge[N << 1];
inline void add(long long u, long long v) {
  edge[++cnt] = {v, head[u]};
  head[u] = cnt;
}
inline void dfs(long long u, long long fa) {
  f[u] = sz[u] = 1;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    f[u] += g[v];
    g[u] += f[v];
  }
  ans = (ans + (f[u] - g[u]) * (n - sz[u] + 1) % mod * a[u] % mod) % mod;
  ans = (ans + (sz[u] - 1) * a[u] % mod) % mod;
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    ans = (ans + (g[v] - f[v]) * (sz[u] - sz[v] - 1) % mod * a[u] % mod) % mod;
  }
}
long long uf[N], ug[N];
inline void work(long long u, long long fa) {
  for (long long i = head[u]; i; i = edge[i].next) {
    long long v = edge[i].to;
    if (v == fa) continue;
    uf[v] = ug[u] + g[u] - f[v];
    ug[v] = uf[u] + f[u] - g[v];
    ans = (ans + (uf[v] - ug[v]) * sz[v] % mod * a[v] % mod) % mod;
    work(v, u);
  }
}
signed main() {
  n = read();
  long long u, v;
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i < n; i++) {
    u = read(), v = read();
    add(u, v), add(v, u);
  }
  dfs(1, 0);
  work(1, 0);
  cout << (ans + mod) % mod;
  return 0;
  ;
}

```