#include <bits/stdc++.h>
using namespace std;
void qmax(long long &x, long long y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline int read() {
  char s;
  int k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(int x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 100;
int n, a[maxn], X, Y;
int to[maxn * 2], ne[maxn * 2], po[maxn], id;
long long ans, dp[maxn][2], f[maxn][2];
void dfs(int x, int fa) {
  dp[x][0] = a[x];
  f[x][0] = 1;
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != fa) {
      dfs(to[i], x);
      dp[x][0] += dp[*(to + i)][1] + a[x] * f[*(to + i)][1] % mod;
      dp[x][0] %= mod;
      f[x][0] += f[*(to + i)][1];
      f[x][0] %= mod;
      dp[x][1] += dp[*(to + i)][0] - a[x] * f[*(to + i)][0] % mod;
      dp[x][1] %= mod;
      f[x][1] += f[*(to + i)][0];
      f[x][1] %= mod;
    }
  long long sum = dp[x][0];
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != fa) {
      int y = to[i];
      sum += dp[y][0] * (f[x][1] - f[y][0]) % mod +
             (dp[x][1] - dp[y][0] + f[y][0] * a[x] % mod) * f[y][0] % mod;
      sum += dp[y][1] * (f[x][0] - f[y][1]) % mod +
             (dp[x][0] - dp[y][1] - f[y][1] * a[x] % mod) * f[y][1] % mod;
      sum %= mod;
    }
  ans += sum;
  ans %= mod;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i < n; i++) {
    X = read(), Y = read();
    id++, to[id] = Y, ne[id] = po[X], po[X] = id, id++, to[id] = X,
          ne[id] = po[Y], po[Y] = id;
  }
  dfs(1, 0);
  ans = (ans % mod + mod) % mod;
  printf("%lld\n", ans);
  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>
using namespace std;
void qmax(long long &x, long long y) {
  if (x < y) x = y;
}
void qmin(int &x, int y) {
  if (x > y) x = y;
}
inline int read() {
  char s;
  int k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) {
    k = k * 10 + (s ^ '0');
    s = getchar();
  }
  return k * base;
}
inline void write(int x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    printf("0");
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 100;
int n, a[maxn], X, Y;
int to[maxn * 2], ne[maxn * 2], po[maxn], id;
long long ans, dp[maxn][2], f[maxn][2];
void dfs(int x, int fa) {
  dp[x][0] = a[x];
  f[x][0] = 1;
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != fa) {
      dfs(to[i], x);
      dp[x][0] += dp[*(to + i)][1] + a[x] * f[*(to + i)][1] % mod;
      dp[x][0] %= mod;
      f[x][0] += f[*(to + i)][1];
      f[x][0] %= mod;
      dp[x][1] += dp[*(to + i)][0] - a[x] * f[*(to + i)][0] % mod;
      dp[x][1] %= mod;
      f[x][1] += f[*(to + i)][0];
      f[x][1] %= mod;
    }
  long long sum = dp[x][0];
  for (int i = po[x]; i; i = ne[i])
    if (to[i] != fa) {
      int y = to[i];
      sum += dp[y][0] * (f[x][1] - f[y][0]) % mod +
             (dp[x][1] - dp[y][0] + f[y][0] * a[x] % mod) * f[y][0] % mod;
      sum += dp[y][1] * (f[x][0] - f[y][1]) % mod +
             (dp[x][0] - dp[y][1] - f[y][1] * a[x] % mod) * f[y][1] % mod;
      sum %= mod;
    }
  ans += sum;
  ans %= mod;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i < n; i++) {
    X = read(), Y = read();
    id++, to[id] = Y, ne[id] = po[X], po[X] = id, id++, to[id] = X,
          ne[id] = po[Y], po[Y] = id;
  }
  dfs(1, 0);
  ans = (ans % mod + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

```