#include <bits/stdc++.h>
using namespace std;
const int N = 200010, P = 1000000007;
long long sum[2], ans, tsum[2];
int root, total;
int a[N], h[N], siz[N], cnt[2], tcnt[2];
bool v[N];
vector<int> g[N];
void dfs1(int x, int tp) {
  siz[x] = 1;
  h[x] = 0;
  for (auto p : g[x])
    if (p != tp && !v[p]) {
      dfs1(p, x);
      siz[x] += siz[p];
      h[x] = max(h[x], siz[p]);
    }
  h[x] = max(h[x], total - siz[x]);
  if (!root || h[x] < h[root]) root = x;
}
void dfs2(int x, int tp, int len, long long val) {
  (tsum[len] += val) %= P;
  siz[x] = 1;
  tcnt[len]++;
  for (auto p : g[x])
    if (p != tp && !v[p]) {
      dfs2(p, x, len ^ 1, (val + (len ? (P - 1ll) : 1ll) * a[p] % P) % P);
      siz[x] += siz[p];
    }
}
void solve(int x) {
  v[x] = 1;
  sum[0] = sum[1] = cnt[0] = cnt[1] = 0;
  (ans += a[x]) %= P;
  for (auto p : g[x])
    if (!v[p]) {
      tsum[0] = tsum[1] = tcnt[0] = tcnt[1] = 0;
      dfs2(p, x, 1, a[p]);
      ans += 2ll * tcnt[0] * cnt[0] % P * a[x] % P +
             1ll * (P - 2) * sum[0] % P * tcnt[0] % P + P -
             2ll * tsum[0] * cnt[0] % P;
      ans += 1ll * (P - 2) * tcnt[1] % P * cnt[1] % P * a[x] % P +
             2ll * sum[1] * tcnt[1] % P + 2ll * tsum[1] * cnt[1] % P;
      ans += 1ll * (P - 2) * tsum[0] % P + 2ll * tcnt[0] * a[x] % P;
      ans %= P;
      (sum[0] += tsum[0]) %= P;
      (sum[1] += tsum[1]) %= P;
      cnt[0] += tcnt[0];
      cnt[1] += tcnt[1];
    }
  for (auto p : g[x])
    if (!v[p]) {
      root = 0, total = siz[p];
      dfs1(p, p);
      solve(root);
    }
}
int main() {
  int n;
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]), a[i] = (a[i] + P) % P;
  int x, y;
  for (int i = (2); i <= (n); i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  total = n, root = 0;
  dfs1(1, 1);
  solve(root);
  cout << ans << endl;
  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 int N = 200010, P = 1000000007;
long long sum[2], ans, tsum[2];
int root, total;
int a[N], h[N], siz[N], cnt[2], tcnt[2];
bool v[N];
vector<int> g[N];
void dfs1(int x, int tp) {
  siz[x] = 1;
  h[x] = 0;
  for (auto p : g[x])
    if (p != tp && !v[p]) {
      dfs1(p, x);
      siz[x] += siz[p];
      h[x] = max(h[x], siz[p]);
    }
  h[x] = max(h[x], total - siz[x]);
  if (!root || h[x] < h[root]) root = x;
}
void dfs2(int x, int tp, int len, long long val) {
  (tsum[len] += val) %= P;
  siz[x] = 1;
  tcnt[len]++;
  for (auto p : g[x])
    if (p != tp && !v[p]) {
      dfs2(p, x, len ^ 1, (val + (len ? (P - 1ll) : 1ll) * a[p] % P) % P);
      siz[x] += siz[p];
    }
}
void solve(int x) {
  v[x] = 1;
  sum[0] = sum[1] = cnt[0] = cnt[1] = 0;
  (ans += a[x]) %= P;
  for (auto p : g[x])
    if (!v[p]) {
      tsum[0] = tsum[1] = tcnt[0] = tcnt[1] = 0;
      dfs2(p, x, 1, a[p]);
      ans += 2ll * tcnt[0] * cnt[0] % P * a[x] % P +
             1ll * (P - 2) * sum[0] % P * tcnt[0] % P + P -
             2ll * tsum[0] * cnt[0] % P;
      ans += 1ll * (P - 2) * tcnt[1] % P * cnt[1] % P * a[x] % P +
             2ll * sum[1] * tcnt[1] % P + 2ll * tsum[1] * cnt[1] % P;
      ans += 1ll * (P - 2) * tsum[0] % P + 2ll * tcnt[0] * a[x] % P;
      ans %= P;
      (sum[0] += tsum[0]) %= P;
      (sum[1] += tsum[1]) %= P;
      cnt[0] += tcnt[0];
      cnt[1] += tcnt[1];
    }
  for (auto p : g[x])
    if (!v[p]) {
      root = 0, total = siz[p];
      dfs1(p, p);
      solve(root);
    }
}
int main() {
  int n;
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%d", &a[i]), a[i] = (a[i] + P) % P;
  int x, y;
  for (int i = (2); i <= (n); i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  total = n, root = 0;
  dfs1(1, 1);
  solve(root);
  cout << ans << endl;
  return 0;
}

```