#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
const int INF = 2139062143;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = 0;
  } while (ch > '9' || ch < '0');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  x = f ? x : -x;
}
template <typename T>
void write(T x) {
  if (x < 0) x = ~x + 1, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 2e5 + 7;
const long long mod = 1e9 + 7;
struct EDGE {
  long long to, nxt;
} edge[N << 1];
long long n, E, ans, a[N], sz[N], head[N], f[N][2], g[N][2];
inline void addedge(long long u, long long v) {
  edge[++E].to = v;
  edge[E].nxt = head[u];
  head[u] = E;
}
inline void dfs(long long u, long long fa) {
  sz[u] = 1;
  f[u][0] = 1;
  f[u][1] = 0;
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    f[u][0] = (f[u][0] + f[v][1]) % mod;
    f[u][1] = (f[u][1] + f[v][0]) % mod;
  }
}
inline void sfd(long long u, long long fa) {
  g[u][0] = f[u][0];
  g[u][1] = f[u][1];
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    f[u][1] = (f[u][1] - f[v][0] + mod) % mod;
    f[u][0] = (f[u][0] - f[v][1] + mod) % mod;
    f[v][1] = (f[v][1] + f[u][0]) % mod;
    f[v][0] = (f[v][0] + f[u][1]) % mod;
    sfd(v, u);
    f[v][1] = (f[v][1] - f[u][0] + mod) % mod;
    f[v][0] = (f[v][0] - f[u][1] + mod) % mod;
    f[u][1] = (f[u][1] + f[v][0]) % mod;
    f[u][0] = (f[u][0] + f[v][1]) % mod;
  }
}
inline void solve(long long u, long long fa) {
  ans = (ans +
         1ll * a[u] * sz[u] % mod * ((g[u][0] - f[u][0] + mod) % mod) % mod -
         1ll * a[u] % mod * sz[u] % mod * ((g[u][1] - f[u][1] + mod) % mod) %
             mod +
         mod) %
        mod;
  ans = (ans % mod + mod) % mod;
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    solve(v, u);
    ans = (ans + 1ll * f[v][1] * (n - sz[v]) % mod * a[u] % mod -
           1ll * f[v][0] * (n - sz[v]) % mod * a[u] % mod + mod) %
          mod;
    ans = (ans % mod + mod) % mod;
  }
}
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++)
    read(a[i]), a[i] = (a[i] % mod + mod) % mod;
  for (long long i = 1, u, v; i < n; i++) {
    read(u);
    read(v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0);
  sfd(1, 0);
  for (long long u = 1; u <= n; u++) ans = (ans + 1ll * n * a[u] % mod) % mod;
  solve(1, 0);
  printf("%lld\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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
const int INF = 2139062143;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = 0;
  } while (ch > '9' || ch < '0');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  x = f ? x : -x;
}
template <typename T>
void write(T x) {
  if (x < 0) x = ~x + 1, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const long long N = 2e5 + 7;
const long long mod = 1e9 + 7;
struct EDGE {
  long long to, nxt;
} edge[N << 1];
long long n, E, ans, a[N], sz[N], head[N], f[N][2], g[N][2];
inline void addedge(long long u, long long v) {
  edge[++E].to = v;
  edge[E].nxt = head[u];
  head[u] = E;
}
inline void dfs(long long u, long long fa) {
  sz[u] = 1;
  f[u][0] = 1;
  f[u][1] = 0;
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
    f[u][0] = (f[u][0] + f[v][1]) % mod;
    f[u][1] = (f[u][1] + f[v][0]) % mod;
  }
}
inline void sfd(long long u, long long fa) {
  g[u][0] = f[u][0];
  g[u][1] = f[u][1];
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    f[u][1] = (f[u][1] - f[v][0] + mod) % mod;
    f[u][0] = (f[u][0] - f[v][1] + mod) % mod;
    f[v][1] = (f[v][1] + f[u][0]) % mod;
    f[v][0] = (f[v][0] + f[u][1]) % mod;
    sfd(v, u);
    f[v][1] = (f[v][1] - f[u][0] + mod) % mod;
    f[v][0] = (f[v][0] - f[u][1] + mod) % mod;
    f[u][1] = (f[u][1] + f[v][0]) % mod;
    f[u][0] = (f[u][0] + f[v][1]) % mod;
  }
}
inline void solve(long long u, long long fa) {
  ans = (ans +
         1ll * a[u] * sz[u] % mod * ((g[u][0] - f[u][0] + mod) % mod) % mod -
         1ll * a[u] % mod * sz[u] % mod * ((g[u][1] - f[u][1] + mod) % mod) %
             mod +
         mod) %
        mod;
  ans = (ans % mod + mod) % mod;
  for (long long i = head[u]; i; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (v == fa) continue;
    solve(v, u);
    ans = (ans + 1ll * f[v][1] * (n - sz[v]) % mod * a[u] % mod -
           1ll * f[v][0] * (n - sz[v]) % mod * a[u] % mod + mod) %
          mod;
    ans = (ans % mod + mod) % mod;
  }
}
signed main() {
  read(n);
  for (long long i = 1; i <= n; i++)
    read(a[i]), a[i] = (a[i] % mod + mod) % mod;
  for (long long i = 1, u, v; i < n; i++) {
    read(u);
    read(v);
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0);
  sfd(1, 0);
  for (long long u = 1; u <= n; u++) ans = (ans + 1ll * n * a[u] % mod) % mod;
  solve(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```