#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 200005;
const long long MOD = 1e9 + 7;
long long n, ans;
long long a[MAXN];
long long dp1[MAXN][2], dp2[MAXN][2];
vector<long long> dis[MAXN][2];
long long Read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
void Put1(long long x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
void Put(long long x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
long long head[MAXN], tot;
struct edge {
  long long v, nxt;
} e[MAXN << 1];
void Add_Edge(long long x, long long y) {
  e[++tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void Add_Double_Edge(long long x, long long y) {
  Add_Edge(x, y);
  Add_Edge(y, x);
}
void dfs1(long long x, long long fa) {
  dp1[x][1] = 1;
  for (long long i = head[x]; i; i = e[i].nxt) {
    if (e[i].v == fa) continue;
    dfs1(e[i].v, x);
    dp1[x][0] += dp1[e[i].v][1];
    dis[x][0].push_back(dp1[e[i].v][1]);
    dp1[x][1] += dp1[e[i].v][0];
    dis[x][1].push_back(dp1[e[i].v][0]);
  }
}
void dfs2(long long x, long long fa) {
  dp2[x][1] = 1;
  if (x != 1) {
    dp2[x][0] += dp2[fa][1] + dp1[fa][1] - 1 - dp1[x][0];
    dp2[x][1] += dp2[fa][0] + dp1[fa][0] - dp1[x][1];
  }
  dis[x][0].push_back(dp2[x][0]);
  dis[x][1].push_back(dp2[x][1] - 1);
  for (long long i = head[x]; i; i = e[i].nxt) {
    if (e[i].v == fa) continue;
    dfs2(e[i].v, x);
  }
}
signed main() {
  n = Read();
  for (long long i = 1; i <= n; ++i) a[i] = Read();
  for (long long i = 1; i < n; ++i) Add_Double_Edge(Read(), Read());
  dfs1(1, 0);
  dfs2(1, 0);
  for (long long i = 1; i <= n; ++i) {
    long long dz = 0;
    for (long long j = 0, len = dis[i][0].size(); j < len; ++j) {
      dz = (dz - 1ll * dis[i][0][j] *
                     ((dp1[i][0] + dp2[i][0] - dis[i][0][j]) % MOD) % MOD *
                     a[i] % MOD) %
           MOD;
    }
    for (long long j = 0, len = dis[i][1].size(); j < len; ++j) {
      dz = (dz + 1ll * dis[i][1][j] *
                     ((dp1[i][1] + dp2[i][1] - 2 - dis[i][1][j]) % MOD) % MOD *
                     a[i] % MOD) %
           MOD;
    }
    dz = (dz + 1ll * a[i] * ((dp1[i][1] + dp2[i][1] - 2) * 2 + 1)) % MOD;
    ans = (ans + dz) % MOD;
  }
  Put((ans + MOD) % MOD);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long MAXN = 200005;
const long long MOD = 1e9 + 7;
long long n, ans;
long long a[MAXN];
long long dp1[MAXN][2], dp2[MAXN][2];
vector<long long> dis[MAXN][2];
long long Read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x * 10) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
void Put1(long long x) {
  if (x > 9) Put1(x / 10);
  putchar(x % 10 ^ 48);
}
void Put(long long x, char c = -1) {
  if (x < 0) putchar('-'), x = -x;
  Put1(x);
  if (c >= 0) putchar(c);
}
template <typename T>
T Max(T x, T y) {
  return x > y ? x : y;
}
template <typename T>
T Min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
T Abs(T x) {
  return x < 0 ? -x : x;
}
long long head[MAXN], tot;
struct edge {
  long long v, nxt;
} e[MAXN << 1];
void Add_Edge(long long x, long long y) {
  e[++tot].v = y;
  e[tot].nxt = head[x];
  head[x] = tot;
}
void Add_Double_Edge(long long x, long long y) {
  Add_Edge(x, y);
  Add_Edge(y, x);
}
void dfs1(long long x, long long fa) {
  dp1[x][1] = 1;
  for (long long i = head[x]; i; i = e[i].nxt) {
    if (e[i].v == fa) continue;
    dfs1(e[i].v, x);
    dp1[x][0] += dp1[e[i].v][1];
    dis[x][0].push_back(dp1[e[i].v][1]);
    dp1[x][1] += dp1[e[i].v][0];
    dis[x][1].push_back(dp1[e[i].v][0]);
  }
}
void dfs2(long long x, long long fa) {
  dp2[x][1] = 1;
  if (x != 1) {
    dp2[x][0] += dp2[fa][1] + dp1[fa][1] - 1 - dp1[x][0];
    dp2[x][1] += dp2[fa][0] + dp1[fa][0] - dp1[x][1];
  }
  dis[x][0].push_back(dp2[x][0]);
  dis[x][1].push_back(dp2[x][1] - 1);
  for (long long i = head[x]; i; i = e[i].nxt) {
    if (e[i].v == fa) continue;
    dfs2(e[i].v, x);
  }
}
signed main() {
  n = Read();
  for (long long i = 1; i <= n; ++i) a[i] = Read();
  for (long long i = 1; i < n; ++i) Add_Double_Edge(Read(), Read());
  dfs1(1, 0);
  dfs2(1, 0);
  for (long long i = 1; i <= n; ++i) {
    long long dz = 0;
    for (long long j = 0, len = dis[i][0].size(); j < len; ++j) {
      dz = (dz - 1ll * dis[i][0][j] *
                     ((dp1[i][0] + dp2[i][0] - dis[i][0][j]) % MOD) % MOD *
                     a[i] % MOD) %
           MOD;
    }
    for (long long j = 0, len = dis[i][1].size(); j < len; ++j) {
      dz = (dz + 1ll * dis[i][1][j] *
                     ((dp1[i][1] + dp2[i][1] - 2 - dis[i][1][j]) % MOD) % MOD *
                     a[i] % MOD) %
           MOD;
    }
    dz = (dz + 1ll * a[i] * ((dp1[i][1] + dp2[i][1] - 2) * 2 + 1)) % MOD;
    ans = (ans + dz) % MOD;
  }
  Put((ans + MOD) % MOD);
  return 0;
}

```