#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const long long ms = 1e9 + 7;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
struct Edg {
  int nxt, poi;
} e[N << 1];
long long f[N][2], g[N][2], ans = 0, a[N];
long long size[N], l, first[N], n;
void addedge(int u, int v) {
  l++;
  e[l].nxt = first[u];
  e[l].poi = v;
  first[u] = l;
}
void dfs(int u, int fa) {
  size[u] = 0;
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    dfs(v, u);
    f[u][1] = (f[u][1] + f[v][0]) % ms;
    f[u][0] = (f[u][0] + f[v][1]) % ms;
    size[u] += size[v];
  }
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    ans =
        (ans + (size[u] - size[v]) * (f[v][1] - f[v][0]) % ms * a[u] % ms) % ms;
  }
  ans = (ans + (n - size[u]) * (f[u][0] - f[u][1]) % ms * a[u] % ms) % ms;
  f[u][0]++;
  if (f[u][0] >= ms) f[u][0] -= ms;
  size[u]++;
}
void dfs2(int u, int fa) {
  if (u != 1) ans = (ans + size[u] * (g[u][0] - g[u][1]) % ms * a[u] % ms) % ms;
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    g[v][0] = (f[u][1] - f[v][0] + g[u][1]) % ms;
    g[v][1] = (f[u][0] - f[v][1] + g[u][0]) % ms;
    dfs2(v, u);
  }
  ans = (ans + n * a[u]) % ms;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    read(u);
    read(v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 1);
  dfs2(1, 1);
  ans += ms;
  if (ans >= ms) ans -= ms;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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 int N = 200005;
const long long ms = 1e9 + 7;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
struct Edg {
  int nxt, poi;
} e[N << 1];
long long f[N][2], g[N][2], ans = 0, a[N];
long long size[N], l, first[N], n;
void addedge(int u, int v) {
  l++;
  e[l].nxt = first[u];
  e[l].poi = v;
  first[u] = l;
}
void dfs(int u, int fa) {
  size[u] = 0;
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    dfs(v, u);
    f[u][1] = (f[u][1] + f[v][0]) % ms;
    f[u][0] = (f[u][0] + f[v][1]) % ms;
    size[u] += size[v];
  }
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    ans =
        (ans + (size[u] - size[v]) * (f[v][1] - f[v][0]) % ms * a[u] % ms) % ms;
  }
  ans = (ans + (n - size[u]) * (f[u][0] - f[u][1]) % ms * a[u] % ms) % ms;
  f[u][0]++;
  if (f[u][0] >= ms) f[u][0] -= ms;
  size[u]++;
}
void dfs2(int u, int fa) {
  if (u != 1) ans = (ans + size[u] * (g[u][0] - g[u][1]) % ms * a[u] % ms) % ms;
  for (int p = first[u]; p; p = e[p].nxt) {
    int v = e[p].poi;
    if (v == fa) continue;
    g[v][0] = (f[u][1] - f[v][0] + g[u][1]) % ms;
    g[v][1] = (f[u][0] - f[v][1] + g[u][0]) % ms;
    dfs2(v, u);
  }
  ans = (ans + n * a[u]) % ms;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    read(u);
    read(v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 1);
  dfs2(1, 1);
  ans += ms;
  if (ans >= ms) ans -= ms;
  printf("%I64d\n", ans);
  return 0;
}

```