#include <bits/stdc++.h>
using namespace std;
struct ppl {
  int64_t sm[2], sz[2];
  ppl() { sm[0] = sm[1] = sz[0] = sz[1] = 0LL; }
  ppl(pair<int64_t, int64_t> a, pair<int64_t, int64_t> b) {
    sm[0] = a.first, sm[1] = a.second;
    sz[0] = b.first, sz[1] = b.second;
  }
};
int n;
int num[200005];
int64_t ans;
vector<int> adj[200005];
const int md = 1e9 + 7;
void merge(ppl &a, const ppl b) {
  a.sz[0] += b.sz[0];
  a.sz[1] += b.sz[1];
  a.sm[0] += b.sm[0];
  a.sm[1] += b.sm[1];
  a.sz[0] %= md;
  a.sz[1] %= md;
  a.sm[0] %= md;
  a.sm[1] %= md;
}
ppl dfs(int st, int pr) {
  ppl cp;
  for (auto x : adj[st]) {
    if (x == pr) {
      continue;
    }
    auto p = dfs(x, st);
    ans += (p.sm[0]) * (cp.sz[0]) << 1;
    ans += (p.sm[1]) * (cp.sz[1]) << 1;
    ans %= md;
    ans += (cp.sm[0]) * (p.sz[0]) << 1;
    ans += (cp.sm[1]) * (p.sz[1]) << 1;
    ans %= md;
    ans -= num[st] * 1LL * (((cp.sz[0] * p.sz[0]) << 1) % md);
    ans += num[st] * 1LL * (((cp.sz[1] * p.sz[1]) << 1) % md);
    ans %= md;
    merge(cp, p);
  }
  ans += num[st];
  cp.sm[0] -= num[st] * cp.sz[0];
  cp.sm[1] += num[st] * cp.sz[1];
  cp.sm[0] %= md;
  cp.sm[1] %= md;
  ans += cp.sm[1] << 1;
  ans %= md;
  cp.sz[1]++;
  cp.sm[1] += num[st];
  swap(cp.sm[0], cp.sm[1]);
  swap(cp.sz[0], cp.sz[1]);
  return cp;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", (ans + md) % md);
}

### 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;
struct ppl {
  int64_t sm[2], sz[2];
  ppl() { sm[0] = sm[1] = sz[0] = sz[1] = 0LL; }
  ppl(pair<int64_t, int64_t> a, pair<int64_t, int64_t> b) {
    sm[0] = a.first, sm[1] = a.second;
    sz[0] = b.first, sz[1] = b.second;
  }
};
int n;
int num[200005];
int64_t ans;
vector<int> adj[200005];
const int md = 1e9 + 7;
void merge(ppl &a, const ppl b) {
  a.sz[0] += b.sz[0];
  a.sz[1] += b.sz[1];
  a.sm[0] += b.sm[0];
  a.sm[1] += b.sm[1];
  a.sz[0] %= md;
  a.sz[1] %= md;
  a.sm[0] %= md;
  a.sm[1] %= md;
}
ppl dfs(int st, int pr) {
  ppl cp;
  for (auto x : adj[st]) {
    if (x == pr) {
      continue;
    }
    auto p = dfs(x, st);
    ans += (p.sm[0]) * (cp.sz[0]) << 1;
    ans += (p.sm[1]) * (cp.sz[1]) << 1;
    ans %= md;
    ans += (cp.sm[0]) * (p.sz[0]) << 1;
    ans += (cp.sm[1]) * (p.sz[1]) << 1;
    ans %= md;
    ans -= num[st] * 1LL * (((cp.sz[0] * p.sz[0]) << 1) % md);
    ans += num[st] * 1LL * (((cp.sz[1] * p.sz[1]) << 1) % md);
    ans %= md;
    merge(cp, p);
  }
  ans += num[st];
  cp.sm[0] -= num[st] * cp.sz[0];
  cp.sm[1] += num[st] * cp.sz[1];
  cp.sm[0] %= md;
  cp.sm[1] %= md;
  ans += cp.sm[1] << 1;
  ans %= md;
  cp.sz[1]++;
  cp.sm[1] += num[st];
  swap(cp.sm[0], cp.sm[1]);
  swap(cp.sz[0], cp.sz[1]);
  return cp;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  printf("%lld\n", (ans + md) % md);
}

```