#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MOD = 1000000007;
int n;
int v[MAXN];
vector<int> adj[MAXN];
int numdn[MAXN][2];
int numup[MAXN][2];
void dfs1(int at, int par) {
  numdn[at][0] = 0, numdn[at][1] = 1;
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    dfs1(to, at);
    numdn[at][0] = (numdn[at][0] + numdn[to][1]) % MOD;
    numdn[at][1] = (numdn[at][1] + numdn[to][0]) % MOD;
  }
}
void dfs2(int at, int par) {
  numup[at][0] = 0, numup[at][1] = 1;
  if (par != -1) {
    numup[at][0] = ((long long)numup[at][0] + numup[par][1] + numdn[par][1] +
                    MOD - numdn[at][0] + MOD - 1) %
                   MOD;
    numup[at][1] = ((long long)numup[at][1] + numup[par][0] + numdn[par][0] +
                    MOD - numdn[at][1]) %
                   MOD;
  }
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    dfs2(to, at);
  }
}
int dfs3(int at, int par) {
  int ret = 0;
  for (int z = (0); z < (2); ++z) {
    int cur = 0;
    int numtot = ((long long)numdn[at][0] + numdn[at][1] + numup[at][0] +
                  numup[at][1] + MOD - 1) %
                 MOD;
    if (z == 0) {
      cur = (cur + numtot) % MOD;
    }
    if (par != -1) {
      int numdiff =
          ((long long)numtot + MOD - numup[at][0] + MOD - numup[at][1] + 1) %
          MOD;
      int ways = z == 1 ? numup[at][1 - z] : (numup[at][1 - z] + MOD - 1) % MOD;
      cur = (cur + (long long)ways * numdiff) % MOD;
    }
    for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
      int to = adj[at][i];
      if (to == par) continue;
      int numdiff =
          ((long long)numtot + MOD - numdn[to][0] + MOD - numdn[to][1]) % MOD;
      cur = (cur + (long long)numdn[to][z] * numdiff) % MOD;
    }
    int vz = v[at];
    if (z == 1) vz = -vz;
    ret = (ret + (long long)cur * ((vz % MOD) + MOD)) % MOD;
  }
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    ret = (ret + dfs3(to, at)) % MOD;
  }
  return ret;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &v[i]);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs1(0, -1);
  dfs2(0, -1);
  int ans = dfs3(0, -1);
  printf("%d\n", ans);
}
int main() {
  run();
  return 0;
}

### Prompt

Your challenge is to write 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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 200000;
const int MOD = 1000000007;
int n;
int v[MAXN];
vector<int> adj[MAXN];
int numdn[MAXN][2];
int numup[MAXN][2];
void dfs1(int at, int par) {
  numdn[at][0] = 0, numdn[at][1] = 1;
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    dfs1(to, at);
    numdn[at][0] = (numdn[at][0] + numdn[to][1]) % MOD;
    numdn[at][1] = (numdn[at][1] + numdn[to][0]) % MOD;
  }
}
void dfs2(int at, int par) {
  numup[at][0] = 0, numup[at][1] = 1;
  if (par != -1) {
    numup[at][0] = ((long long)numup[at][0] + numup[par][1] + numdn[par][1] +
                    MOD - numdn[at][0] + MOD - 1) %
                   MOD;
    numup[at][1] = ((long long)numup[at][1] + numup[par][0] + numdn[par][0] +
                    MOD - numdn[at][1]) %
                   MOD;
  }
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    dfs2(to, at);
  }
}
int dfs3(int at, int par) {
  int ret = 0;
  for (int z = (0); z < (2); ++z) {
    int cur = 0;
    int numtot = ((long long)numdn[at][0] + numdn[at][1] + numup[at][0] +
                  numup[at][1] + MOD - 1) %
                 MOD;
    if (z == 0) {
      cur = (cur + numtot) % MOD;
    }
    if (par != -1) {
      int numdiff =
          ((long long)numtot + MOD - numup[at][0] + MOD - numup[at][1] + 1) %
          MOD;
      int ways = z == 1 ? numup[at][1 - z] : (numup[at][1 - z] + MOD - 1) % MOD;
      cur = (cur + (long long)ways * numdiff) % MOD;
    }
    for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
      int to = adj[at][i];
      if (to == par) continue;
      int numdiff =
          ((long long)numtot + MOD - numdn[to][0] + MOD - numdn[to][1]) % MOD;
      cur = (cur + (long long)numdn[to][z] * numdiff) % MOD;
    }
    int vz = v[at];
    if (z == 1) vz = -vz;
    ret = (ret + (long long)cur * ((vz % MOD) + MOD)) % MOD;
  }
  for (int i = (0); i < (((int)(adj[at]).size())); ++i) {
    int to = adj[at][i];
    if (to == par) continue;
    ret = (ret + dfs3(to, at)) % MOD;
  }
  return ret;
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d", &v[i]);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs1(0, -1);
  dfs2(0, -1);
  int ans = dfs3(0, -1);
  printf("%d\n", ans);
}
int main() {
  run();
  return 0;
}

```