#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
vector<long long> vec[200005];
map<long long, long long> mp;
long long d[200005], tmp[2], son[200005], ans[200005], v[200005], n;
void dfs1(long long u, long long fa) {
  d[u] = d[fa] + 1;
  tmp[d[u] % 2]++;
  son[u] = 1;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    dfs1(v, u);
    son[u] += son[v];
  }
}
void add(long long u, long long fa) {
  mp[d[u] % 2]++;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    add(v, u);
  }
}
void dfs2(long long u, long long fa) {
  long long bc = 0;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    if (!bc || son[bc] < son[v]) bc = v;
  }
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa || v == bc) continue;
    dfs2(v, u);
    ans[u] = ans[u] + mp[d[u] % 2] * (son[1] - son[v]);
    ans[u] = ans[u] - mp[(d[u] % 2 + 1) % 2] * (son[1] - son[v]);
    mp.clear();
  }
  if (bc) dfs2(bc, u);
  ans[u] = ans[u] + mp[d[u] % 2] * (son[1] - son[bc]);
  ans[u] = ans[u] - mp[(d[u] % 2 + 1) % 2] * (son[1] - son[bc]);
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa || v == bc) continue;
    add(v, u);
  }
  mp[d[u] % 2]++;
  ans[u] = ans[u] + son[1] - 1;
  ans[u] = ans[u] + (tmp[d[u] % 2] - mp[d[u] % 2]) * son[u];
  ans[u] = ans[u] - (tmp[(d[u] % 2 + 1) % 2] - mp[(d[u] % 2 + 1) % 2]) * son[u];
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &v[i]);
  }
  for (long long i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  memset(son, 0, sizeof(son));
  memset(ans, 0, sizeof(ans));
  memset(d, 0, sizeof(d));
  dfs1(1, 0);
  dfs2(1, 0);
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    while (ans[i] < 0) ans[i] += mod;
    ans[i] %= mod;
    while (v[i] < 0) v[i] += mod;
    v[i] %= mod;
    sum = (sum + ans[i] * v[i]) % mod;
    sum = (sum + v[i] + mod) % mod;
  }
  cout << sum << endl;
  return 0;
}

### 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;
const long long mod = 1e9 + 7;
vector<long long> vec[200005];
map<long long, long long> mp;
long long d[200005], tmp[2], son[200005], ans[200005], v[200005], n;
void dfs1(long long u, long long fa) {
  d[u] = d[fa] + 1;
  tmp[d[u] % 2]++;
  son[u] = 1;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    dfs1(v, u);
    son[u] += son[v];
  }
}
void add(long long u, long long fa) {
  mp[d[u] % 2]++;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    add(v, u);
  }
}
void dfs2(long long u, long long fa) {
  long long bc = 0;
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa) continue;
    if (!bc || son[bc] < son[v]) bc = v;
  }
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa || v == bc) continue;
    dfs2(v, u);
    ans[u] = ans[u] + mp[d[u] % 2] * (son[1] - son[v]);
    ans[u] = ans[u] - mp[(d[u] % 2 + 1) % 2] * (son[1] - son[v]);
    mp.clear();
  }
  if (bc) dfs2(bc, u);
  ans[u] = ans[u] + mp[d[u] % 2] * (son[1] - son[bc]);
  ans[u] = ans[u] - mp[(d[u] % 2 + 1) % 2] * (son[1] - son[bc]);
  for (long long i = 0; i < vec[u].size(); i++) {
    long long v = vec[u][i];
    if (v == fa || v == bc) continue;
    add(v, u);
  }
  mp[d[u] % 2]++;
  ans[u] = ans[u] + son[1] - 1;
  ans[u] = ans[u] + (tmp[d[u] % 2] - mp[d[u] % 2]) * son[u];
  ans[u] = ans[u] - (tmp[(d[u] % 2 + 1) % 2] - mp[(d[u] % 2 + 1) % 2]) * son[u];
}
int main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &v[i]);
  }
  for (long long i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  memset(son, 0, sizeof(son));
  memset(ans, 0, sizeof(ans));
  memset(d, 0, sizeof(d));
  dfs1(1, 0);
  dfs2(1, 0);
  long long sum = 0;
  for (long long i = 1; i <= n; i++) {
    while (ans[i] < 0) ans[i] += mod;
    ans[i] %= mod;
    while (v[i] < 0) v[i] += mod;
    v[i] %= mod;
    sum = (sum + ans[i] * v[i]) % mod;
    sum = (sum + v[i] + mod) % mod;
  }
  cout << sum << endl;
  return 0;
}

```