#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
const long long mod = 1e9 + 7;
long long n, val[maxn], siz[maxn], root, head[maxn], cnt, vis[maxn], son[maxn],
    ans, ji, ou, nowji, nowou, dis[maxn], mx, tot[2], sum[2];
struct edge {
  long long to, v, next;
} e[maxn << 1];
void add(long long a, long long b) {
  e[cnt].to = b;
  e[cnt].next = head[a];
  head[a] = cnt++;
}
void getroot(long long u, long long pre) {
  siz[u] = 1;
  son[u] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to] || e[i].to == pre) continue;
    getroot(e[i].to, u);
    siz[u] += siz[e[i].to];
    son[u] = max(son[u], siz[e[i].to]);
  }
  son[u] = max(son[u], mx - siz[u]);
  if (son[u] < son[root]) root = u;
}
void getans(long long u, long long pre, long long o) {
  dis[u] = -dis[pre] + val[u];
  dis[u] = (dis[u] + mod) % mod;
  if (o == 1) {
    ans = (ans + dis[u] * 2 % mod) % mod;
    ans =
        (ans + 2 * ((sum[1] % mod + tot[1] * dis[u] % mod) % mod) % mod) % mod;
  } else {
    ans =
        (ans + 2 * ((sum[0] % mod + tot[0] * dis[u] % mod) % mod) % mod) % mod;
  }
  for (long long i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getans(e[i].to, u, o ^ 1);
  }
}
void getdis(long long u, long long pre, long long o) {
  dis[u] = -dis[pre] + val[u];
  dis[u] = (dis[u] % mod + mod) % mod;
  tot[o]++;
  if (tot[o] >= mod) tot[o] -= mod;
  sum[o] = (sum[o] + dis[u]) % mod;
  for (long long i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getdis(e[i].to, u, o ^ 1);
  }
}
void calc(long long u) {
  tot[0] = tot[1] = sum[0] = sum[1] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to]) continue;
    dis[u] = val[u];
    getans(e[i].to, u, 0);
    dis[u] = 0;
    getdis(e[i].to, u, 0);
  }
  ans = (ans + val[u]) % mod;
}
void dfs(long long u) {
  vis[u] = 1;
  calc(u);
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to]) continue;
    root = 0;
    mx = siz[e[i].to];
    getroot(e[i].to, 0);
    dfs(root);
  }
}
int main() {
  scanf("%lld", &n);
  long long a, b, c;
  ans = 0;
  cnt = 1;
  for (long long i = 1; i <= n; ++i)
    scanf("%lld", &val[i]), val[i] = (val[i] % mod + mod) % mod;
  for (long long i = 1; i < n; ++i) {
    scanf("%lld%lld", &a, &b);
    add(a, b);
    add(b, a);
  }
  root = 0;
  siz[0] = son[0] = mx = n + 1;
  getroot(1, 0);
  dfs(root);
  printf("%lld\n", (ans % mod + mod) % mod);
}

### 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;
const int maxn = 200005;
const long long mod = 1e9 + 7;
long long n, val[maxn], siz[maxn], root, head[maxn], cnt, vis[maxn], son[maxn],
    ans, ji, ou, nowji, nowou, dis[maxn], mx, tot[2], sum[2];
struct edge {
  long long to, v, next;
} e[maxn << 1];
void add(long long a, long long b) {
  e[cnt].to = b;
  e[cnt].next = head[a];
  head[a] = cnt++;
}
void getroot(long long u, long long pre) {
  siz[u] = 1;
  son[u] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to] || e[i].to == pre) continue;
    getroot(e[i].to, u);
    siz[u] += siz[e[i].to];
    son[u] = max(son[u], siz[e[i].to]);
  }
  son[u] = max(son[u], mx - siz[u]);
  if (son[u] < son[root]) root = u;
}
void getans(long long u, long long pre, long long o) {
  dis[u] = -dis[pre] + val[u];
  dis[u] = (dis[u] + mod) % mod;
  if (o == 1) {
    ans = (ans + dis[u] * 2 % mod) % mod;
    ans =
        (ans + 2 * ((sum[1] % mod + tot[1] * dis[u] % mod) % mod) % mod) % mod;
  } else {
    ans =
        (ans + 2 * ((sum[0] % mod + tot[0] * dis[u] % mod) % mod) % mod) % mod;
  }
  for (long long i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getans(e[i].to, u, o ^ 1);
  }
}
void getdis(long long u, long long pre, long long o) {
  dis[u] = -dis[pre] + val[u];
  dis[u] = (dis[u] % mod + mod) % mod;
  tot[o]++;
  if (tot[o] >= mod) tot[o] -= mod;
  sum[o] = (sum[o] + dis[u]) % mod;
  for (long long i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getdis(e[i].to, u, o ^ 1);
  }
}
void calc(long long u) {
  tot[0] = tot[1] = sum[0] = sum[1] = 0;
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to]) continue;
    dis[u] = val[u];
    getans(e[i].to, u, 0);
    dis[u] = 0;
    getdis(e[i].to, u, 0);
  }
  ans = (ans + val[u]) % mod;
}
void dfs(long long u) {
  vis[u] = 1;
  calc(u);
  for (long long i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to]) continue;
    root = 0;
    mx = siz[e[i].to];
    getroot(e[i].to, 0);
    dfs(root);
  }
}
int main() {
  scanf("%lld", &n);
  long long a, b, c;
  ans = 0;
  cnt = 1;
  for (long long i = 1; i <= n; ++i)
    scanf("%lld", &val[i]), val[i] = (val[i] % mod + mod) % mod;
  for (long long i = 1; i < n; ++i) {
    scanf("%lld%lld", &a, &b);
    add(a, b);
    add(b, a);
  }
  root = 0;
  siz[0] = son[0] = mx = n + 1;
  getroot(1, 0);
  dfs(root);
  printf("%lld\n", (ans % mod + mod) % mod);
}

```