#include <bits/stdc++.h>
using namespace std;
inline int readint() {
  int a = 0;
  char c = getchar(), f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -f;
  for (; '0' <= c && c <= '9'; c = getchar())
    a = (a << 3) + (a << 1) + (c ^ 48);
  return a * f;
}
inline void writeint(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) writeint(x / 10);
  putchar((x % 10) ^ 48);
}
template <class T>
void getMax(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
void getMin(T& a, T b) {
  if (b < a) a = b;
}
const int MaxN = 200005;
const int Mod = 1e9 + 7;
struct Edge {
  int to, nxt;
  Edge(int T = 0, int N = 0) { to = T, nxt = N; }
} edge[MaxN << 1];
int head[MaxN], cntEdge;
void addEdge(int a, int b) {
  edge[cntEdge] = Edge(b, head[a]);
  head[a] = cntEdge++;
  edge[cntEdge] = Edge(a, head[b]);
  head[b] = cntEdge++;
}
int depth[MaxN], a[MaxN], n;
int cnt[MaxN][2];
int ans;
void dfs(int x, int pre) {
  int xs = 0;
  cnt[x][depth[x] & 1] = 1;
  for (int i = head[x], y; ~i; i = edge[i].nxt) {
    if ((y = edge[i].to) == pre) continue;
    depth[y] = depth[x] + 1, dfs(y, x);
    for (int j = 0; j < 2; ++j) cnt[x][j] += cnt[y][j];
    xs += 1ll * (n - cnt[y][0] - cnt[y][1]) *
          (cnt[y][depth[x] & 1] - cnt[y][(depth[x] & 1) ^ 1]) % Mod;
    xs = (xs % Mod + Mod) % Mod;
  }
  ans = (ans + 1ll * xs * a[x]) % Mod;
}
int main() {
  n = readint();
  for (int i = 1; i <= n; ++i) {
    a[i] = (readint() + Mod) % Mod;
    head[i] = -1;
  }
  for (int i = 1; i < n; ++i) addEdge(readint(), readint());
  dfs(1, -1);
  for (int i = 1; i <= n; ++i) {
    int d = depth[i] & 1;
    int xs = 1ll * (cnt[i][0] + cnt[i][1]) *
             (cnt[1][d] + cnt[i][d ^ 1] - cnt[1][d ^ 1] - cnt[i][d]) % Mod;
    xs = (1ll * xs % Mod + Mod + n) % Mod;
    ans = (ans + 1ll * xs * a[i]) % Mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate a cpp solution to 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;
inline int readint() {
  int a = 0;
  char c = getchar(), f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -f;
  for (; '0' <= c && c <= '9'; c = getchar())
    a = (a << 3) + (a << 1) + (c ^ 48);
  return a * f;
}
inline void writeint(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) writeint(x / 10);
  putchar((x % 10) ^ 48);
}
template <class T>
void getMax(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
void getMin(T& a, T b) {
  if (b < a) a = b;
}
const int MaxN = 200005;
const int Mod = 1e9 + 7;
struct Edge {
  int to, nxt;
  Edge(int T = 0, int N = 0) { to = T, nxt = N; }
} edge[MaxN << 1];
int head[MaxN], cntEdge;
void addEdge(int a, int b) {
  edge[cntEdge] = Edge(b, head[a]);
  head[a] = cntEdge++;
  edge[cntEdge] = Edge(a, head[b]);
  head[b] = cntEdge++;
}
int depth[MaxN], a[MaxN], n;
int cnt[MaxN][2];
int ans;
void dfs(int x, int pre) {
  int xs = 0;
  cnt[x][depth[x] & 1] = 1;
  for (int i = head[x], y; ~i; i = edge[i].nxt) {
    if ((y = edge[i].to) == pre) continue;
    depth[y] = depth[x] + 1, dfs(y, x);
    for (int j = 0; j < 2; ++j) cnt[x][j] += cnt[y][j];
    xs += 1ll * (n - cnt[y][0] - cnt[y][1]) *
          (cnt[y][depth[x] & 1] - cnt[y][(depth[x] & 1) ^ 1]) % Mod;
    xs = (xs % Mod + Mod) % Mod;
  }
  ans = (ans + 1ll * xs * a[x]) % Mod;
}
int main() {
  n = readint();
  for (int i = 1; i <= n; ++i) {
    a[i] = (readint() + Mod) % Mod;
    head[i] = -1;
  }
  for (int i = 1; i < n; ++i) addEdge(readint(), readint());
  dfs(1, -1);
  for (int i = 1; i <= n; ++i) {
    int d = depth[i] & 1;
    int xs = 1ll * (cnt[i][0] + cnt[i][1]) *
             (cnt[1][d] + cnt[i][d ^ 1] - cnt[1][d ^ 1] - cnt[i][d]) % Mod;
    xs = (1ll * xs % Mod + Mod + n) % Mod;
    ans = (ans + 1ll * xs * a[i]) % Mod;
  }
  printf("%d\n", ans);
  return 0;
}

```