#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
const long long mod = 1e9 + 7;
const int inf = 1e9;
long long val[N], op[2], ct[2], ans;
int n, _size, mx, rt;
int cur, h[N], nex[N << 1], to[N << 1], siz[N], vis[N];
void add_edge(int x, int y) {
  to[++cur] = y;
  nex[cur] = h[x];
  h[x] = cur;
}
void getrt(int u, int fa) {
  siz[u] = 1;
  int num = 0;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    getrt(v, u);
    siz[u] += siz[v];
    if (siz[v] > num) num = siz[v];
  }
  if (_size - siz[u] > num) num = _size - siz[u];
  if (num < mx) mx = num, rt = u;
}
void cal(int u, int fa, int dep, long long nowsum, long long add) {
  if (dep) {
    nowsum = (nowsum + val[u]) % mod;
    ans = (ans + (ct[dep] * (nowsum - add + mod) % mod + op[dep]) % mod) % mod;
  } else {
    nowsum = (nowsum - val[u] + mod) % mod;
    ans =
        (ans + (ct[dep] * (-nowsum + add + mod) % mod - op[dep] + mod) % mod) %
        mod;
  }
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    cal(v, u, dep ^ 1, nowsum, add);
  }
}
void ins(int u, int fa, int dep, long long s) {
  if (dep)
    s = (s + val[u]) % mod;
  else
    s = (s - val[u] + mod) % mod;
  ct[dep]++;
  op[dep] = (op[dep] + s) % mod;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    ins(v, u, dep ^ 1, s);
  }
}
void divide(int u) {
  vis[u] = 1;
  op[0] = 0;
  op[1] = 0;
  ct[0] = 1, ct[1] = 0;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (vis[v]) continue;
    cal(v, u, 1, 0, val[u]);
    ins(v, u, 1, 0);
  }
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (vis[v]) continue;
    _size = siz[v];
    mx = inf;
    getrt(v, u);
    divide(rt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%lld", &val[i]), val[i] = (val[i] + mod) % mod;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  _size = n;
  mx = inf;
  getrt(1, 0);
  divide(rt);
  ans = (ans * 2ll) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + val[i]) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int N = 2e5 + 100;
const long long mod = 1e9 + 7;
const int inf = 1e9;
long long val[N], op[2], ct[2], ans;
int n, _size, mx, rt;
int cur, h[N], nex[N << 1], to[N << 1], siz[N], vis[N];
void add_edge(int x, int y) {
  to[++cur] = y;
  nex[cur] = h[x];
  h[x] = cur;
}
void getrt(int u, int fa) {
  siz[u] = 1;
  int num = 0;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    getrt(v, u);
    siz[u] += siz[v];
    if (siz[v] > num) num = siz[v];
  }
  if (_size - siz[u] > num) num = _size - siz[u];
  if (num < mx) mx = num, rt = u;
}
void cal(int u, int fa, int dep, long long nowsum, long long add) {
  if (dep) {
    nowsum = (nowsum + val[u]) % mod;
    ans = (ans + (ct[dep] * (nowsum - add + mod) % mod + op[dep]) % mod) % mod;
  } else {
    nowsum = (nowsum - val[u] + mod) % mod;
    ans =
        (ans + (ct[dep] * (-nowsum + add + mod) % mod - op[dep] + mod) % mod) %
        mod;
  }
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    cal(v, u, dep ^ 1, nowsum, add);
  }
}
void ins(int u, int fa, int dep, long long s) {
  if (dep)
    s = (s + val[u]) % mod;
  else
    s = (s - val[u] + mod) % mod;
  ct[dep]++;
  op[dep] = (op[dep] + s) % mod;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (v == fa || vis[v]) continue;
    ins(v, u, dep ^ 1, s);
  }
}
void divide(int u) {
  vis[u] = 1;
  op[0] = 0;
  op[1] = 0;
  ct[0] = 1, ct[1] = 0;
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (vis[v]) continue;
    cal(v, u, 1, 0, val[u]);
    ins(v, u, 1, 0);
  }
  for (int i = h[u]; i; i = nex[i]) {
    int v = to[i];
    if (vis[v]) continue;
    _size = siz[v];
    mx = inf;
    getrt(v, u);
    divide(rt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%lld", &val[i]), val[i] = (val[i] + mod) % mod;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  _size = n;
  mx = inf;
  getrt(1, 0);
  divide(rt);
  ans = (ans * 2ll) % mod;
  for (int i = 1; i <= n; i++) ans = (ans + val[i]) % mod;
  printf("%lld\n", ans);
  return 0;
}

```