#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 4e5 + 100;
const long long MOD = 1e9 + 7;
struct Edge {
  long long from, to, nxt;
} edge[MAXN];
long long eidx;
long long head[MAXN];
void addedge(long long from, long long to) {
  edge[eidx].from = from;
  edge[eidx].to = to;
  edge[eidx].nxt = head[from];
  head[from] = eidx++;
}
void einit() {
  eidx = 0;
  memset(head, -1, sizeof head);
}
long long val[MAXN];
long long o[MAXN], e[MAXN], lev[MAXN];
long long ans;
bool vis[MAXN];
void dfs(long long u, long long h, long long fa) {
  lev[u] = h;
  vis[u] = 1;
  o[u] = e[u] = 0;
  for (long long i = head[u]; i != -1; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (vis[v]) continue;
    dfs(v, h + 1, u);
    long long ta =
        2 * (o[u] * e[v] % MOD - o[v] * e[u] % MOD) % MOD * val[u] % MOD;
    ta = (ta % MOD + MOD) % MOD;
    if (u == 1) {
    }
    ans = (ans + ta) % MOD;
    o[u] += e[v];
    e[u] += o[v];
  }
  o[u]++;
}
void init() { memset(vis, 0, sizeof vis); }
int main() {
  long long n;
  {
    cin >> n;
    einit();
    init();
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", &val[i]);
    }
    for (int i = 1; i < n; ++i) {
      long long p1, p2;
      scanf("%lld%lld", &p1, &p2);
      addedge(p1, p2);
      addedge(p2, p1);
    }
    ans = 0;
    dfs(1, 1, -1);
    for (int i = 1; i <= n; ++i) {
      long long upo, upe, ta;
      if (lev[i] % 2) {
        upo = o[1] - o[i] + 1;
        upe = e[1] - e[i];
      } else {
        upo = e[1] - o[i] + 1;
        upe = o[1] - e[i];
      }
      ta = 2 * (upo * o[i] % MOD - upe * e[i] % MOD) % MOD * val[i] % MOD;
      ta -= val[i];
      ta = (ta % MOD + MOD) % MOD;
      ans = (ans + ta) % MOD;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 long long MAXN = 4e5 + 100;
const long long MOD = 1e9 + 7;
struct Edge {
  long long from, to, nxt;
} edge[MAXN];
long long eidx;
long long head[MAXN];
void addedge(long long from, long long to) {
  edge[eidx].from = from;
  edge[eidx].to = to;
  edge[eidx].nxt = head[from];
  head[from] = eidx++;
}
void einit() {
  eidx = 0;
  memset(head, -1, sizeof head);
}
long long val[MAXN];
long long o[MAXN], e[MAXN], lev[MAXN];
long long ans;
bool vis[MAXN];
void dfs(long long u, long long h, long long fa) {
  lev[u] = h;
  vis[u] = 1;
  o[u] = e[u] = 0;
  for (long long i = head[u]; i != -1; i = edge[i].nxt) {
    long long v = edge[i].to;
    if (vis[v]) continue;
    dfs(v, h + 1, u);
    long long ta =
        2 * (o[u] * e[v] % MOD - o[v] * e[u] % MOD) % MOD * val[u] % MOD;
    ta = (ta % MOD + MOD) % MOD;
    if (u == 1) {
    }
    ans = (ans + ta) % MOD;
    o[u] += e[v];
    e[u] += o[v];
  }
  o[u]++;
}
void init() { memset(vis, 0, sizeof vis); }
int main() {
  long long n;
  {
    cin >> n;
    einit();
    init();
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", &val[i]);
    }
    for (int i = 1; i < n; ++i) {
      long long p1, p2;
      scanf("%lld%lld", &p1, &p2);
      addedge(p1, p2);
      addedge(p2, p1);
    }
    ans = 0;
    dfs(1, 1, -1);
    for (int i = 1; i <= n; ++i) {
      long long upo, upe, ta;
      if (lev[i] % 2) {
        upo = o[1] - o[i] + 1;
        upe = e[1] - e[i];
      } else {
        upo = e[1] - o[i] + 1;
        upe = o[1] - e[i];
      }
      ta = 2 * (upo * o[i] % MOD - upe * e[i] % MOD) % MOD * val[i] % MOD;
      ta -= val[i];
      ta = (ta % MOD + MOD) % MOD;
      ans = (ans + ta) % MOD;
    }
    cout << ans << endl;
  }
  return 0;
}

```