#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
const int N = 2e5 + 10, mod = 1e9 + 7;
int dp1[N][2], dp2[N][2], sz[N], head[N], val[N], ecnt, n;
long long ans;
struct ed {
  int to, next;
} e[N << 1];
void ad(int x, int y) {
  e[++ecnt] = (ed){y, head[x]};
  head[x] = ecnt;
  e[++ecnt] = (ed){x, head[y]};
  head[y] = ecnt;
}
void dfs1(int u, int last) {
  dp1[u][1] = 0, dp1[u][0] = 1, sz[u] = 1;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    dfs1(v, u), dp1[u][1] += dp1[v][0], dp1[u][0] += dp1[v][1];
    ans = (ans +
           val[u] * (long long)(dp1[v][1] - dp1[v][0]) % mod * (n - sz[v])) %
          mod,
    sz[u] += sz[v];
  }
  ans = (ans + val[u] * (long long)n) % mod;
}
void dfs2(int u, int last) {
  if (last)
    dp2[u][1] = dp1[last][0] - dp1[u][1] + dp2[last][0],
    dp2[u][0] = dp1[last][1] - dp1[u][0] + dp2[last][1],
    ans =
        (ans + sz[u] * (long long)(dp2[u][0] - dp2[u][1]) % mod * val[u]) % mod;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    dfs2(v, u);
  }
}
int main() {
  int x, y;
  read(n);
  for (int i = 1; i <= n; i++) read(val[i]);
  for (int i = 1; i < n; i++) read(x), read(y), ad(x, y);
  dfs1(1, 0), dfs2(1, 0), printf("%lld", (ans + 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;
template <class T>
inline void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 - '0' + c, c = getchar();
  x *= f;
}
const int N = 2e5 + 10, mod = 1e9 + 7;
int dp1[N][2], dp2[N][2], sz[N], head[N], val[N], ecnt, n;
long long ans;
struct ed {
  int to, next;
} e[N << 1];
void ad(int x, int y) {
  e[++ecnt] = (ed){y, head[x]};
  head[x] = ecnt;
  e[++ecnt] = (ed){x, head[y]};
  head[y] = ecnt;
}
void dfs1(int u, int last) {
  dp1[u][1] = 0, dp1[u][0] = 1, sz[u] = 1;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    dfs1(v, u), dp1[u][1] += dp1[v][0], dp1[u][0] += dp1[v][1];
    ans = (ans +
           val[u] * (long long)(dp1[v][1] - dp1[v][0]) % mod * (n - sz[v])) %
          mod,
    sz[u] += sz[v];
  }
  ans = (ans + val[u] * (long long)n) % mod;
}
void dfs2(int u, int last) {
  if (last)
    dp2[u][1] = dp1[last][0] - dp1[u][1] + dp2[last][0],
    dp2[u][0] = dp1[last][1] - dp1[u][0] + dp2[last][1],
    ans =
        (ans + sz[u] * (long long)(dp2[u][0] - dp2[u][1]) % mod * val[u]) % mod;
  for (int k = head[u]; k; k = e[k].next) {
    int v = e[k].to;
    if (v == last) continue;
    dfs2(v, u);
  }
}
int main() {
  int x, y;
  read(n);
  for (int i = 1; i <= n; i++) read(val[i]);
  for (int i = 1; i < n; i++) read(x), read(y), ad(x, y);
  dfs1(1, 0), dfs2(1, 0), printf("%lld", (ans + mod) % mod);
  return 0;
}

```