#include <bits/stdc++.h>
using namespace std;
namespace zzc {
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
int n, cnt = 0, root;
int head[maxn], siz[maxn], m1[maxn], m2[maxn], w[maxn], id[maxn], fa[maxn],
    dep[maxn];
int um1[maxn], um2[maxn];
long long ans = 0;
struct edge {
  int to, nxt;
} e[maxn << 1];
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
int dfs(int u, int f) {
  m2[u] = 1;
  siz[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == f) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    siz[u] += dfs(v, u);
    m1[u] += m2[v];
    m2[u] += m1[v];
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == f) continue;
    ans = (ans + 1LL * (m1[v] - m2[v]) % mod * w[u] % mod *
                     (siz[u] - siz[v] - 1) % mod) %
          mod;
  }
  ans = (ans + 1LL * w[u] * (siz[u] - 1) % mod) % mod;
  ans = (ans +
         1LL * (m2[u] - m1[u]) % mod * w[u] % mod * (n - siz[u] + 1) % mod) %
        mod;
  return siz[u];
}
bool cmp(int a, int b) { return dep[a] < dep[b]; }
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]), id[i] = i;
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  dfs(1, 0);
  sort(id + 1, id + n + 1, cmp);
  for (int i = 2; i <= n; i++) {
    int u = id[i];
    um1[u] = um2[fa[u]] + m2[fa[u]] - m1[u];
    um2[u] = um1[fa[u]] + m1[fa[u]] - m2[u];
    ans =
        (ans + 1LL * (um2[u] - um1[u]) % mod * w[u] % mod * siz[u] % mod) % mod;
  }
  printf("%lld", (ans + mod) % mod);
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
namespace zzc {
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
int n, cnt = 0, root;
int head[maxn], siz[maxn], m1[maxn], m2[maxn], w[maxn], id[maxn], fa[maxn],
    dep[maxn];
int um1[maxn], um2[maxn];
long long ans = 0;
struct edge {
  int to, nxt;
} e[maxn << 1];
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
int dfs(int u, int f) {
  m2[u] = 1;
  siz[u] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == f) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    siz[u] += dfs(v, u);
    m1[u] += m2[v];
    m2[u] += m1[v];
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == f) continue;
    ans = (ans + 1LL * (m1[v] - m2[v]) % mod * w[u] % mod *
                     (siz[u] - siz[v] - 1) % mod) %
          mod;
  }
  ans = (ans + 1LL * w[u] * (siz[u] - 1) % mod) % mod;
  ans = (ans +
         1LL * (m2[u] - m1[u]) % mod * w[u] % mod * (n - siz[u] + 1) % mod) %
        mod;
  return siz[u];
}
bool cmp(int a, int b) { return dep[a] < dep[b]; }
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]), id[i] = i;
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    add(a, b);
    add(b, a);
  }
  dfs(1, 0);
  sort(id + 1, id + n + 1, cmp);
  for (int i = 2; i <= n; i++) {
    int u = id[i];
    um1[u] = um2[fa[u]] + m2[fa[u]] - m1[u];
    um2[u] = um1[fa[u]] + m1[fa[u]] - m2[u];
    ans =
        (ans + 1LL * (um2[u] - um1[u]) % mod * w[u] % mod * siz[u] % mod) % mod;
  }
  printf("%lld", (ans + mod) % mod);
}
}  // namespace zzc
int main() {
  zzc::work();
  return 0;
}

```