#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
inline unsigned int myrand(int n) {
  unsigned int u = rand() % n;
  int c = rand() % 10;
  u ^= u << c;
  u ^= (u << 7) & 263692864;
  u ^= (u << 15) & 402273075;
  u %= n;
  return u;
}
int head[400444], to[400444], nxt[400444], val[400444];
int E;
long long ans = 0;
int cnt[2][400444], son[400444];
void add(int u, int v) {
  nxt[E] = head[u];
  to[E] = v;
  head[u] = E++;
  nxt[E] = head[v];
  to[E] = u;
  head[v] = E++;
}
void dfs(int u, int fa = 0) {
  son[u] = cnt[0][u] = 1;
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    dfs(v, u);
    son[u] += son[v];
    cnt[0][u] += cnt[1][v];
    cnt[1][u] += cnt[0][v];
  }
}
int n;
void solve(int u, int fa = 0, int odd = 0, int even = 1) {
  (ans += (long long)even * son[u] % 1000000007 * val[u] % 1000000007) %=
      1000000007;
  (ans -= (long long)odd * son[u] % 1000000007 * val[u] % 1000000007) %=
      1000000007;
  (ans += (long long)(n - son[u]) * val[u] % 1000000007) %= 1000000007;
  int sum[2] = {0, 0};
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    sum[0] += cnt[0][v];
    sum[1] += cnt[1][v];
    (ans += (long long)cnt[1][v] * (n - son[v]) % 1000000007 * val[u] %
            1000000007) %= 1000000007;
    (ans -= (long long)cnt[0][v] * (n - son[v]) % 1000000007 * val[u] %
            1000000007) %= 1000000007;
  }
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    solve(v, u, even + sum[1] - cnt[1][v], odd + sum[0] - cnt[0][v] + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) gn(val[i]);
  memset(head, -1, sizeof head);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    gn(u, v);
    add(u, v);
  }
  dfs(1);
  solve(1);
  if (ans < 0) ans += 1000000007;
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
inline unsigned int myrand(int n) {
  unsigned int u = rand() % n;
  int c = rand() % 10;
  u ^= u << c;
  u ^= (u << 7) & 263692864;
  u ^= (u << 15) & 402273075;
  u %= n;
  return u;
}
int head[400444], to[400444], nxt[400444], val[400444];
int E;
long long ans = 0;
int cnt[2][400444], son[400444];
void add(int u, int v) {
  nxt[E] = head[u];
  to[E] = v;
  head[u] = E++;
  nxt[E] = head[v];
  to[E] = u;
  head[v] = E++;
}
void dfs(int u, int fa = 0) {
  son[u] = cnt[0][u] = 1;
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    dfs(v, u);
    son[u] += son[v];
    cnt[0][u] += cnt[1][v];
    cnt[1][u] += cnt[0][v];
  }
}
int n;
void solve(int u, int fa = 0, int odd = 0, int even = 1) {
  (ans += (long long)even * son[u] % 1000000007 * val[u] % 1000000007) %=
      1000000007;
  (ans -= (long long)odd * son[u] % 1000000007 * val[u] % 1000000007) %=
      1000000007;
  (ans += (long long)(n - son[u]) * val[u] % 1000000007) %= 1000000007;
  int sum[2] = {0, 0};
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    sum[0] += cnt[0][v];
    sum[1] += cnt[1][v];
    (ans += (long long)cnt[1][v] * (n - son[v]) % 1000000007 * val[u] %
            1000000007) %= 1000000007;
    (ans -= (long long)cnt[0][v] * (n - son[v]) % 1000000007 * val[u] %
            1000000007) %= 1000000007;
  }
  for (int e = head[u], v; ~e; e = nxt[e]) {
    v = to[e];
    if (v == fa) continue;
    solve(v, u, even + sum[1] - cnt[1][v], odd + sum[0] - cnt[0][v] + 1);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) gn(val[i]);
  memset(head, -1, sizeof head);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    gn(u, v);
    add(u, v);
  }
  dfs(1);
  solve(1);
  if (ans < 0) ans += 1000000007;
  cout << ans << endl;
  return 0;
}

```