#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n;
long long ans;
int val[maxn];
int head[maxn], nxt[maxn << 1], to[maxn << 1], edge_tot;
void add_edge(int u, int v) {
  nxt[++edge_tot] = head[u];
  head[u] = edge_tot;
  to[edge_tot] = v;
}
int root;
bool vis[maxn];
int siz[maxn], mxson[maxn];
void ndc_init() { mxson[0] = inf; }
void add(long long &a, long long b) { (a += b) %= mod; }
void del(long long &a, long long b) {
  (a -= b) %= mod;
  if (a < 0) a += mod;
}
void mul(long long &a, long long b) { a = a * b % mod; }
void get_root(int u, int f, int sum) {
  siz[u] = 1;
  mxson[u] = 0;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    get_root(v, u, sum);
    siz[u] += siz[v];
    if (mxson[u] < siz[v]) mxson[u] = siz[v];
  }
  if (mxson[u] < sum - siz[u]) mxson[u] = sum - siz[u];
  if (mxson[u] < mxson[root]) root = u;
}
long long ret[2];
void add(int u, int f, int dep, long long sum, int op) {
  if (dep & 1)
    add(sum, val[u]);
  else
    del(sum, val[u]);
  add(ret[1], op * sum);
  add(ret[0], op * (-sum));
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    add(v, u, dep + 1, sum, op);
  }
}
void calc(int u, int f, long long sum, int dep, int rot_val, int cnt) {
  if (dep & 1)
    add(ans, 1ll * cnt * (sum - rot_val) % mod + ret[1]);
  else
    add(ans, 1ll * cnt * (-sum + rot_val) % mod + ret[0]);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    long long res = sum;
    if (dep & 1)
      del(res, val[v]);
    else
      add(res, val[v]);
    calc(v, u, res, dep + 1, rot_val, cnt);
  }
}
void get_ans(int u, int f, int dep, long long sum) {
  if (dep & 1)
    add(sum, val[u]);
  else
    del(sum, val[u]);
  add(ans, sum);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    get_ans(v, u, dep + 1, sum);
  }
}
void solve(int u = 1, int sum = n) {
  root = 0;
  get_root(u, 0, sum);
  vis[root] = 1;
  get_ans(root, 0, 1, 0);
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, 1);
  }
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, -1);
    int cnt = 0;
    if (siz[root] > siz[v])
      cnt = sum - siz[v];
    else
      cnt = siz[root];
    calc(v, root, val[v], 1, val[root], cnt);
    add(v, root, 1, 0, 1);
  }
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, -1);
  }
  int root_sz = siz[root];
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    solve(v, root_sz > siz[v] ? siz[v] : sum - root_sz);
  }
}
int main() {
  ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  ndc_init();
  solve();
  printf("%lld\n", (ans % mod + mod) % mod);
  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;
const int maxn = 2e5 + 50;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
int n;
long long ans;
int val[maxn];
int head[maxn], nxt[maxn << 1], to[maxn << 1], edge_tot;
void add_edge(int u, int v) {
  nxt[++edge_tot] = head[u];
  head[u] = edge_tot;
  to[edge_tot] = v;
}
int root;
bool vis[maxn];
int siz[maxn], mxson[maxn];
void ndc_init() { mxson[0] = inf; }
void add(long long &a, long long b) { (a += b) %= mod; }
void del(long long &a, long long b) {
  (a -= b) %= mod;
  if (a < 0) a += mod;
}
void mul(long long &a, long long b) { a = a * b % mod; }
void get_root(int u, int f, int sum) {
  siz[u] = 1;
  mxson[u] = 0;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    get_root(v, u, sum);
    siz[u] += siz[v];
    if (mxson[u] < siz[v]) mxson[u] = siz[v];
  }
  if (mxson[u] < sum - siz[u]) mxson[u] = sum - siz[u];
  if (mxson[u] < mxson[root]) root = u;
}
long long ret[2];
void add(int u, int f, int dep, long long sum, int op) {
  if (dep & 1)
    add(sum, val[u]);
  else
    del(sum, val[u]);
  add(ret[1], op * sum);
  add(ret[0], op * (-sum));
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    add(v, u, dep + 1, sum, op);
  }
}
void calc(int u, int f, long long sum, int dep, int rot_val, int cnt) {
  if (dep & 1)
    add(ans, 1ll * cnt * (sum - rot_val) % mod + ret[1]);
  else
    add(ans, 1ll * cnt * (-sum + rot_val) % mod + ret[0]);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    long long res = sum;
    if (dep & 1)
      del(res, val[v]);
    else
      add(res, val[v]);
    calc(v, u, res, dep + 1, rot_val, cnt);
  }
}
void get_ans(int u, int f, int dep, long long sum) {
  if (dep & 1)
    add(sum, val[u]);
  else
    del(sum, val[u]);
  add(ans, sum);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v] || v == f) continue;
    get_ans(v, u, dep + 1, sum);
  }
}
void solve(int u = 1, int sum = n) {
  root = 0;
  get_root(u, 0, sum);
  vis[root] = 1;
  get_ans(root, 0, 1, 0);
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, 1);
  }
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, -1);
    int cnt = 0;
    if (siz[root] > siz[v])
      cnt = sum - siz[v];
    else
      cnt = siz[root];
    calc(v, root, val[v], 1, val[root], cnt);
    add(v, root, 1, 0, 1);
  }
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    add(v, root, 1, 0, -1);
  }
  int root_sz = siz[root];
  for (int i = head[root]; i; i = nxt[i]) {
    int v = to[i];
    if (vis[v]) continue;
    solve(v, root_sz > siz[v] ? siz[v] : sum - root_sz);
  }
}
int main() {
  ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  ndc_init();
  solve();
  printf("%lld\n", (ans % mod + mod) % mod);
  return 0;
}

```