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. 
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long double PI = acos(-1);
const int N = 2e5 + 5;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
long long n, o[N], e[N], vis[N], par[N];
map<pair<int, int>, long long> ee, oo;
long long sz[N];
long long v[N];
long long ans;
vector<int> adj[N];
pair<int, int> dfs(int u) {
  vis[u] = 1;
  for (int v : adj[u])
    if (!vis[v]) {
      par[v] = u;
      pair<int, int> p = dfs(v);
      e[u] = p.second + e[u];
      o[u] = p.first + o[u];
      ee[make_pair(u, v)] = o[v];
      oo[make_pair(u, v)] = e[v];
    }
  o[u]++;
  return {e[u], o[u]};
}
void dfs_p(int u, int p) {
  if (p) {
    ee[make_pair(u, p)] = o[p] - e[u];
    oo[make_pair(u, p)] = e[p] - o[u];
    o[u] += oo[make_pair(u, p)];
    e[u] += ee[make_pair(u, p)];
  }
  vis[u] = 1;
  for (int v : adj[u])
    if (!vis[v]) {
      dfs_p(v, u);
    }
}
long long dfs_t(int u) {
  vis[u] = 1;
  for (int v : adj[u])
    if (!vis[v]) {
      sz[u] += dfs_t(v);
    }
  sz[u]++;
  return sz[u];
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  memset(vis, 0, sizeof vis);
  dfs_p(1, 0);
  memset(vis, 0, sizeof vis);
  dfs_t(1);
  for (int i = 1; i <= n; i++) {
    ans = (ans + (n * v[i])) % MOD;
    long long ev = 0;
    long long od = 0;
    for (int j : adj[i]) {
      long long qt = par[j] == i ? n - sz[j] : sz[i];
      long long todd = ((oo[make_pair(i, j)] * qt) % MOD);
      long long teven = ((ee[make_pair(i, j)] * qt) % MOD);
      ev = (ev + teven) % MOD;
      od = (od + todd) % MOD;
    }
    ans = (ans - ((ev * v[i]) % MOD) + MOD) % MOD;
    ans = (ans + ((od * v[i]) % MOD) + MOD) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
