#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  x = 0;
  char c;
  long long sign = 1;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (!isdigit(c));
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (isdigit(c));
  x *= sign;
}
const long long N = 4e5 + 50;
const long long p = 1e9 + 7;
long long head[N], nxt[N], to[N], tot;
long long n, m;
long long val[N];
inline void add(long long x, long long y) {
  nxt[tot] = head[x];
  to[tot] = y;
  head[x] = tot++;
}
long long rt, nsum, mx[N], size[N];
bool vis[N];
void get_rt(long long x, long long fa) {
  size[x] = mx[x] = 1;
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]] && to[i] != fa) {
      get_rt(to[i], x);
      size[x] += size[to[i]];
      mx[x] = max(mx[x], size[to[i]]);
    }
  mx[x] = max(mx[x], nsum - size[x]);
  if (mx[rt] > mx[x]) rt = x;
}
long long d[2], num[2];
void dfs(long long x, long long fa, long long n, long long w) {
  w %= p;
  d[n] += w;
  d[n] %= p;
  ++num[n];
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]] && to[i] != fa) {
      dfs(to[i], x, n ^ 1, val[to[i]] * (n ? 1 : -1) + w);
    }
}
long long calc1(long long x) {
  d[0] = d[1] = 0;
  num[0] = num[1] = 0;
  dfs(x, x, 0, 0);
  d[1] = -d[1];
  return (d[0] * num[0] * 2 % p + d[1] * num[1] * 2 % p +
          val[x] * (((num[0] * num[0] - num[1] * num[1]) % p + p) % p)) %
         p;
}
long long calc2(long long x, long long w) {
  d[0] = d[1] = 0;
  num[0] = num[1] = 0;
  dfs(x, x, 1, -val[x]);
  d[1] = -d[1];
  return (d[0] * num[0] * 2 % p + d[1] * num[1] * 2 % p +
          w * (((num[0] * num[0] - num[1] * num[1]) % p + p) % p) % p) %
         p;
}
long long ans;
void solve(long long x) {
  vis[x] = 1;
  ans = (ans + calc1(x)) % p;
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]]) {
      ans = (ans - calc2(to[i], val[x])) % p;
      rt = 0;
      nsum = size[to[i]];
      get_rt(to[i], to[i]);
      solve(rt);
    }
}
int main() {
  memset(head, -1, sizeof head);
  read(n);
  for (register long long i = 1; i <= n; ++i) read(val[i]);
  long long u, v;
  for (register long long i = 2; i <= n; ++i) {
    read(u);
    read(v);
    add(u, v);
    add(v, u);
  }
  mx[0] = N;
  nsum = n;
  get_rt(1, 1);
  solve(rt);
  cout << (ans + p) % p << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 void read(T& x) {
  x = 0;
  char c;
  long long sign = 1;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (!isdigit(c));
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (isdigit(c));
  x *= sign;
}
const long long N = 4e5 + 50;
const long long p = 1e9 + 7;
long long head[N], nxt[N], to[N], tot;
long long n, m;
long long val[N];
inline void add(long long x, long long y) {
  nxt[tot] = head[x];
  to[tot] = y;
  head[x] = tot++;
}
long long rt, nsum, mx[N], size[N];
bool vis[N];
void get_rt(long long x, long long fa) {
  size[x] = mx[x] = 1;
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]] && to[i] != fa) {
      get_rt(to[i], x);
      size[x] += size[to[i]];
      mx[x] = max(mx[x], size[to[i]]);
    }
  mx[x] = max(mx[x], nsum - size[x]);
  if (mx[rt] > mx[x]) rt = x;
}
long long d[2], num[2];
void dfs(long long x, long long fa, long long n, long long w) {
  w %= p;
  d[n] += w;
  d[n] %= p;
  ++num[n];
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]] && to[i] != fa) {
      dfs(to[i], x, n ^ 1, val[to[i]] * (n ? 1 : -1) + w);
    }
}
long long calc1(long long x) {
  d[0] = d[1] = 0;
  num[0] = num[1] = 0;
  dfs(x, x, 0, 0);
  d[1] = -d[1];
  return (d[0] * num[0] * 2 % p + d[1] * num[1] * 2 % p +
          val[x] * (((num[0] * num[0] - num[1] * num[1]) % p + p) % p)) %
         p;
}
long long calc2(long long x, long long w) {
  d[0] = d[1] = 0;
  num[0] = num[1] = 0;
  dfs(x, x, 1, -val[x]);
  d[1] = -d[1];
  return (d[0] * num[0] * 2 % p + d[1] * num[1] * 2 % p +
          w * (((num[0] * num[0] - num[1] * num[1]) % p + p) % p) % p) %
         p;
}
long long ans;
void solve(long long x) {
  vis[x] = 1;
  ans = (ans + calc1(x)) % p;
  for (register long long i = head[x]; ~i; i = nxt[i])
    if (!vis[to[i]]) {
      ans = (ans - calc2(to[i], val[x])) % p;
      rt = 0;
      nsum = size[to[i]];
      get_rt(to[i], to[i]);
      solve(rt);
    }
}
int main() {
  memset(head, -1, sizeof head);
  read(n);
  for (register long long i = 1; i <= n; ++i) read(val[i]);
  long long u, v;
  for (register long long i = 2; i <= n; ++i) {
    read(u);
    read(v);
    add(u, v);
    add(v, u);
  }
  mx[0] = N;
  nsum = n;
  get_rt(1, 1);
  solve(rt);
  cout << (ans + p) % p << endl;
  return 0;
}

```