#include <bits/stdc++.h>
using namespace std;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1e6 + 10;
vector<pair<int, int> > e[SIZE];
int eo[SIZE];
long long w[SIZE];
int dp[SIZE][2];
void dfs_eo(int x, int lt, int v) {
  eo[x] = v;
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    dfs_eo(y, x, v ^ 1);
  }
}
long long an = 0;
void dfs1(int x, int lt) {
  dp[x][eo[x]]++;
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    dfs1(y, x);
    for (int j = 0; j < (2); ++j) dp[x][j] += dp[y][j];
  }
}
long long sqr(long long x) { return x * x % MOD; }
void dfs2(int x, int lt, int now[2]) {
  long long mul = sqr(dp[1][eo[x]]) - sqr(dp[1][eo[x] ^ 1]);
  ADD(mul, -sqr(now[eo[x]]) + sqr(now[eo[x] ^ 1]));
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    ADD(mul, -sqr(dp[y][eo[x]]) + sqr(dp[y][eo[x] ^ 1]));
    int nxt[2] = {now[0], now[1]};
    for (int j = 0; j < (2); ++j) nxt[j] += dp[x][j] - dp[y][j];
    dfs2(y, x, nxt);
  }
  ADD(an, mul * w[x]);
}
void add_edge(int x, int y, int v) {
  e[x].push_back(make_pair(y, v));
  e[y].push_back(make_pair(x, v));
}
int main() {
  int n;
  R(n);
  for (int i = (1); i <= (n); ++i) {
    R(w[i]);
  }
  for (int i = (1); i < (n); ++i) {
    int x, y;
    R(x, y);
    add_edge(x, y, 1);
  }
  dfs_eo(1, 1, 0);
  dfs1(1, 1);
  int ker[2] = {};
  dfs2(1, 1, ker);
  W(an);
  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;
template <class T>
void _R(T &x) {
  cin >> x;
}
void _R(int &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const int &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
int MOD = 1e9 + 7;
void ADD(long long &x, long long v) {
  x = (x + v) % MOD;
  if (x < 0) x += MOD;
}
const int SIZE = 1e6 + 10;
vector<pair<int, int> > e[SIZE];
int eo[SIZE];
long long w[SIZE];
int dp[SIZE][2];
void dfs_eo(int x, int lt, int v) {
  eo[x] = v;
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    dfs_eo(y, x, v ^ 1);
  }
}
long long an = 0;
void dfs1(int x, int lt) {
  dp[x][eo[x]]++;
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    dfs1(y, x);
    for (int j = 0; j < (2); ++j) dp[x][j] += dp[y][j];
  }
}
long long sqr(long long x) { return x * x % MOD; }
void dfs2(int x, int lt, int now[2]) {
  long long mul = sqr(dp[1][eo[x]]) - sqr(dp[1][eo[x] ^ 1]);
  ADD(mul, -sqr(now[eo[x]]) + sqr(now[eo[x] ^ 1]));
  for (int i = 0; i < (((int)(e[x]).size())); ++i) {
    int y = e[x][i].first;
    if (y == lt) continue;
    ADD(mul, -sqr(dp[y][eo[x]]) + sqr(dp[y][eo[x] ^ 1]));
    int nxt[2] = {now[0], now[1]};
    for (int j = 0; j < (2); ++j) nxt[j] += dp[x][j] - dp[y][j];
    dfs2(y, x, nxt);
  }
  ADD(an, mul * w[x]);
}
void add_edge(int x, int y, int v) {
  e[x].push_back(make_pair(y, v));
  e[y].push_back(make_pair(x, v));
}
int main() {
  int n;
  R(n);
  for (int i = (1); i <= (n); ++i) {
    R(w[i]);
  }
  for (int i = (1); i < (n); ++i) {
    int x, y;
    R(x, y);
    add_edge(x, y, 1);
  }
  dfs_eo(1, 1, 0);
  dfs1(1, 1);
  int ker[2] = {};
  dfs2(1, 1, ker);
  W(an);
  return 0;
}

```