#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int MAXN = (int)2e5 + 100;
const int MOD = (int)1e9 + 7;
int n;
int a[MAXN];
vi g[MAXN];
bool read() {
  if (!(cin >> n)) {
    return 0;
  }
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
    a[i] %= MOD;
    a[i] += MOD;
    a[i] %= MOD;
  }
  for (int i = 0; i < int(n); ++i) {
    g[i].clear();
  }
  for (int i = 0; i < int(n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  return 1;
}
int cnt[2][MAXN], h[MAXN];
int sz[MAXN];
void dfs(int v, int p = -1, int ch = 0) {
  sz[v] = 1;
  h[v] = ch;
  cnt[0][v] = cnt[1][v] = 0;
  cnt[ch][v]++;
  for (int i = 0; i < int((g[v]).size()); i++) {
    if (g[v][i] == p) {
      swap(g[v][i], g[v].back());
      g[v].pop_back();
      i--;
      continue;
    }
    int to = g[v][i];
    dfs(to, v, ch ^ 1);
    sz[v] += sz[to];
    for (int j = 0; j < int(2); ++j) {
      cnt[j][v] += cnt[j][to];
    }
  }
}
void add(int &x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) {
    x += MOD;
  }
  return x;
}
int mul(int x, int y) { return 1ll * x * y % MOD; }
int ans;
void calc(int v) {
  for (auto to : g[v]) {
    calc(to);
    int ps = cnt[h[v]][to], ms = cnt[h[v] ^ 1][to];
    int mult = sub(ps, ms);
    add(ans, mul(mul(mult, a[v]), n - sz[to]));
  }
  int ps = cnt[h[v]][0] - cnt[h[v]][v],
      ms = cnt[h[v] ^ 1][0] - cnt[h[v] ^ 1][v];
  int mult = sub(ps, ms);
  add(ans, mul(mul(mult, a[v]), sz[v]));
  add(ans, mul(a[v], n));
}
int solve() {
  ;
  ;
  dfs(0);
  ans = 0;
  calc(0);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  while (read()) {
    cout << solve() << '\n';
  }
}

### Prompt

Please create a solution in CPP 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;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
const int MAXN = (int)2e5 + 100;
const int MOD = (int)1e9 + 7;
int n;
int a[MAXN];
vi g[MAXN];
bool read() {
  if (!(cin >> n)) {
    return 0;
  }
  for (int i = 0; i < int(n); ++i) {
    cin >> a[i];
    a[i] %= MOD;
    a[i] += MOD;
    a[i] %= MOD;
  }
  for (int i = 0; i < int(n); ++i) {
    g[i].clear();
  }
  for (int i = 0; i < int(n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  return 1;
}
int cnt[2][MAXN], h[MAXN];
int sz[MAXN];
void dfs(int v, int p = -1, int ch = 0) {
  sz[v] = 1;
  h[v] = ch;
  cnt[0][v] = cnt[1][v] = 0;
  cnt[ch][v]++;
  for (int i = 0; i < int((g[v]).size()); i++) {
    if (g[v][i] == p) {
      swap(g[v][i], g[v].back());
      g[v].pop_back();
      i--;
      continue;
    }
    int to = g[v][i];
    dfs(to, v, ch ^ 1);
    sz[v] += sz[to];
    for (int j = 0; j < int(2); ++j) {
      cnt[j][v] += cnt[j][to];
    }
  }
}
void add(int &x, int y) {
  x += y;
  if (x >= MOD) {
    x -= MOD;
  }
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) {
    x += MOD;
  }
  return x;
}
int mul(int x, int y) { return 1ll * x * y % MOD; }
int ans;
void calc(int v) {
  for (auto to : g[v]) {
    calc(to);
    int ps = cnt[h[v]][to], ms = cnt[h[v] ^ 1][to];
    int mult = sub(ps, ms);
    add(ans, mul(mul(mult, a[v]), n - sz[to]));
  }
  int ps = cnt[h[v]][0] - cnt[h[v]][v],
      ms = cnt[h[v] ^ 1][0] - cnt[h[v] ^ 1][v];
  int mult = sub(ps, ms);
  add(ans, mul(mul(mult, a[v]), sz[v]));
  add(ans, mul(a[v], n));
}
int solve() {
  ;
  ;
  dfs(0);
  ans = 0;
  calc(0);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  while (read()) {
    cout << solve() << '\n';
  }
}

```