#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long n, a[N], parity[2], val_parity[2], child[N], pref[N], h[N];
bool visited[N];
vector<long long> g[N];
long long ans = 0;
void input() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    scanf("%lld %lld", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(visited, 0, sizeof(visited));
}
void dfs(long long u, long long p) {
  child[u] = 1;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) {
      dfs(v, u);
      child[u] += child[v];
    }
  }
}
long long findcen(long long u, long long p, long long sz) {
  for (long long v : g[u]) {
    if (v != p && !visited[v] && child[v] >= sz / 2) return findcen(v, u, sz);
  }
  return u;
}
void redfs(long long u, long long p) {
  for (long long v : g[u]) {
    if (v != p && !visited[v]) {
      h[v] = h[u] + 1;
      if (h[v] & 1)
        pref[v] = (pref[u] - a[v]) % mod;
      else
        pref[v] = (pref[u] + a[v]) % mod;
      redfs(v, u);
    }
  }
}
void get(long long u, long long p, long long root) {
  long long high = h[u] & 1;
  long long dist = (pref[u] - a[root]) % mod;
  long long cur = (parity[high] * dist + val_parity[high]) % mod;
  if (high)
    ans = (ans - cur) % mod;
  else
    ans = (ans + cur) % mod;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) get(v, u, root);
  }
}
void update(long long u, long long p) {
  long long high = h[u] & 1;
  val_parity[high] = (val_parity[high] + pref[u]) % mod;
  parity[high]++;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) update(v, u);
  }
}
void centroid(long long root) {
  dfs(root, root);
  long long cen = findcen(root, root, child[root]);
  visited[cen] = true;
  h[cen] = 0;
  pref[cen] = a[cen];
  redfs(cen, cen);
  parity[0] = 1;
  parity[1] = 0;
  val_parity[0] = a[cen];
  val_parity[1] = 0;
  for (long long v : g[cen]) {
    if (!visited[v]) {
      get(v, cen, cen);
      update(v, cen);
    }
  }
  for (long long v : g[cen])
    if (!visited[v]) centroid(v);
}
void solve() {
  centroid(1);
  ans = (ans * 2) % mod;
  for (long long i = 1; i <= n; i++) ans = (ans + a[i]) % mod;
  printf("%lld", (ans + mod) % mod);
}
signed main() {
  input();
  solve();
  return 0;
}

### Prompt

Please formulate 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;
const long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long n, a[N], parity[2], val_parity[2], child[N], pref[N], h[N];
bool visited[N];
vector<long long> g[N];
long long ans = 0;
void input() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n - 1; i++) {
    long long u, v;
    scanf("%lld %lld", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(visited, 0, sizeof(visited));
}
void dfs(long long u, long long p) {
  child[u] = 1;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) {
      dfs(v, u);
      child[u] += child[v];
    }
  }
}
long long findcen(long long u, long long p, long long sz) {
  for (long long v : g[u]) {
    if (v != p && !visited[v] && child[v] >= sz / 2) return findcen(v, u, sz);
  }
  return u;
}
void redfs(long long u, long long p) {
  for (long long v : g[u]) {
    if (v != p && !visited[v]) {
      h[v] = h[u] + 1;
      if (h[v] & 1)
        pref[v] = (pref[u] - a[v]) % mod;
      else
        pref[v] = (pref[u] + a[v]) % mod;
      redfs(v, u);
    }
  }
}
void get(long long u, long long p, long long root) {
  long long high = h[u] & 1;
  long long dist = (pref[u] - a[root]) % mod;
  long long cur = (parity[high] * dist + val_parity[high]) % mod;
  if (high)
    ans = (ans - cur) % mod;
  else
    ans = (ans + cur) % mod;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) get(v, u, root);
  }
}
void update(long long u, long long p) {
  long long high = h[u] & 1;
  val_parity[high] = (val_parity[high] + pref[u]) % mod;
  parity[high]++;
  for (long long v : g[u]) {
    if (v != p && !visited[v]) update(v, u);
  }
}
void centroid(long long root) {
  dfs(root, root);
  long long cen = findcen(root, root, child[root]);
  visited[cen] = true;
  h[cen] = 0;
  pref[cen] = a[cen];
  redfs(cen, cen);
  parity[0] = 1;
  parity[1] = 0;
  val_parity[0] = a[cen];
  val_parity[1] = 0;
  for (long long v : g[cen]) {
    if (!visited[v]) {
      get(v, cen, cen);
      update(v, cen);
    }
  }
  for (long long v : g[cen])
    if (!visited[v]) centroid(v);
}
void solve() {
  centroid(1);
  ans = (ans * 2) % mod;
  for (long long i = 1; i <= n; i++) ans = (ans + a[i]) % mod;
  printf("%lld", (ans + mod) % mod);
}
signed main() {
  input();
  solve();
  return 0;
}

```