#include <bits/stdc++.h>
using namespace std;
const long long N = 200004;
vector<long long> g[N];
long long S[N], E[N], O[N], A[N];
long long ans = 0;
inline long long mul(long long a, long long b) { return (a * b) % 1000000007; }
inline long long power_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    b = b / 2;
    a = mul(a, a);
  }
  return ans;
}
inline long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
void dfs(long long x, long long p, long long d) {
  if (d % 2 == 0)
    E[x]++;
  else
    O[x]++;
  S[x]++;
  for (long long y : g[x]) {
    if (y == p) continue;
    dfs(y, x, d + 1);
    E[x] += E[y];
    O[x] += O[y];
    S[x] += S[y];
  }
}
void dfs2(long long x, long long p, long long d) {
  long long tvl = 0;
  for (long long y : g[x]) {
    if (y == p) {
      long long evenUp = E[1] - E[x];
      long long oddUp = O[1] - O[x];
      if (d % 2 == 1) {
        long long vl =
            sub(mul(A[x], mul(oddUp, S[x])), mul(A[x], mul(evenUp, S[x])));
        tvl = add(tvl, vl);
      } else {
        long long vl =
            sub(mul(A[x], mul(evenUp, S[x])), mul(A[x], mul(oddUp, S[x])));
        tvl = add(tvl, vl);
      }
      continue;
    }
    long long evenDown = E[y];
    long long oddDown = O[y];
    if (d % 2 == 1) {
      long long vl = sub(mul(A[x], mul(oddDown, S[1] - S[y])),
                         mul(A[x], mul(evenDown, S[1] - S[y])));
      tvl = add(tvl, vl);
    } else {
      long long vl = sub(mul(A[x], mul(evenDown, S[1] - S[y])),
                         mul(A[x], mul(oddDown, S[1] - S[y])));
      tvl = add(tvl, vl);
    }
    dfs2(y, x, d + 1);
  }
  tvl = add(tvl, mul(A[x], S[1]));
  ans = add(ans, tvl);
}
void solve() {
  long long n, u, v;
  cin >> n;
  for (__typeof((n)) i = (1); i <= (n); i++) {
    cin >> A[i];
    if (A[i] < 0) A[i] = 1000000007 + A[i];
  }
  for (__typeof((n - 1)) i = (1); i <= (n - 1); i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1, 0);
  dfs2(1, 1, 0);
  cout << ans << '\n';
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const long long N = 200004;
vector<long long> g[N];
long long S[N], E[N], O[N], A[N];
long long ans = 0;
inline long long mul(long long a, long long b) { return (a * b) % 1000000007; }
inline long long power_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = mul(ans, a);
    b = b / 2;
    a = mul(a, a);
  }
  return ans;
}
inline long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
void dfs(long long x, long long p, long long d) {
  if (d % 2 == 0)
    E[x]++;
  else
    O[x]++;
  S[x]++;
  for (long long y : g[x]) {
    if (y == p) continue;
    dfs(y, x, d + 1);
    E[x] += E[y];
    O[x] += O[y];
    S[x] += S[y];
  }
}
void dfs2(long long x, long long p, long long d) {
  long long tvl = 0;
  for (long long y : g[x]) {
    if (y == p) {
      long long evenUp = E[1] - E[x];
      long long oddUp = O[1] - O[x];
      if (d % 2 == 1) {
        long long vl =
            sub(mul(A[x], mul(oddUp, S[x])), mul(A[x], mul(evenUp, S[x])));
        tvl = add(tvl, vl);
      } else {
        long long vl =
            sub(mul(A[x], mul(evenUp, S[x])), mul(A[x], mul(oddUp, S[x])));
        tvl = add(tvl, vl);
      }
      continue;
    }
    long long evenDown = E[y];
    long long oddDown = O[y];
    if (d % 2 == 1) {
      long long vl = sub(mul(A[x], mul(oddDown, S[1] - S[y])),
                         mul(A[x], mul(evenDown, S[1] - S[y])));
      tvl = add(tvl, vl);
    } else {
      long long vl = sub(mul(A[x], mul(evenDown, S[1] - S[y])),
                         mul(A[x], mul(oddDown, S[1] - S[y])));
      tvl = add(tvl, vl);
    }
    dfs2(y, x, d + 1);
  }
  tvl = add(tvl, mul(A[x], S[1]));
  ans = add(ans, tvl);
}
void solve() {
  long long n, u, v;
  cin >> n;
  for (__typeof((n)) i = (1); i <= (n); i++) {
    cin >> A[i];
    if (A[i] < 0) A[i] = 1000000007 + A[i];
  }
  for (__typeof((n - 1)) i = (1); i <= (n - 1); i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1, 0);
  dfs2(1, 1, 0);
  cout << ans << '\n';
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) solve();
  return 0;
}

```