#include <bits/stdc++.h>
using namespace std;
const int M = 2e5 + 5, P = 1e9 + 7;
int n, val[M];
int head[M], asdf;
struct edge {
  int to, nxt;
} G[M * 2];
void add_edge(int a, int b) {
  G[++asdf].to = b;
  G[asdf].nxt = head[a];
  head[a] = asdf;
}
int Md(int x) { return (x % P + P) % P; }
bool mark[M];
int sz[M], son[M], que_len, que[M];
void get_root(int x, int f) {
  sz[x] = 1;
  son[x] = 0;
  que[++que_len] = x;
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_root(y, x);
    sz[x] += sz[y];
    if (sz[y] > son[x]) son[x] = sz[y];
  }
}
int ans, sum[2], tot[2];
int dis[M], cnt[M];
void get_ans(int x, int f) {
  cnt[x] = cnt[f] * (-1);
  dis[x] = Md(val[x] - dis[f]);
  if (cnt[x] == 1) {
    ans = Md(ans + 2ll * dis[x] % P);
    ans = Md(1ll * dis[x] * tot[1] % P + ans);
    ans = Md(-sum[1] + ans);
    ans = Md(1ll * dis[x] * tot[1] % P + ans);
    ans = Md(-sum[1] + ans);
  } else {
    ans = Md(1ll * dis[x] * tot[0] % P + ans);
    ans = Md(sum[0] + ans);
    ans = Md(1ll * dis[x] * tot[0] % P + ans);
    ans = Md(sum[0] + ans);
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_ans(y, x);
  }
}
void get_dis(int x, int f) {
  cnt[x] = cnt[f] * (-1);
  dis[x] = Md(dis[f] + val[x] * cnt[x]);
  if (cnt[x] == 1) {
    sum[0] = Md(sum[0] + dis[x]);
    tot[0]++;
    if (tot[0] >= P) tot[0] -= P;
  } else {
    sum[1] = Md(sum[1] + dis[x]);
    tot[1]++;
    if (tot[1] >= P) tot[1] -= P;
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_dis(y, x);
  }
}
void solve(int x) {
  que_len = 0;
  get_root(x, x);
  for (int i = 1; i <= que_len; i++) {
    son[que[i]] = max(son[que[i]], que_len - sz[que[i]]);
    if (son[que[i]] < son[x]) x = que[i];
  }
  mark[x] = 1;
  sum[0] = sum[1] = 0;
  tot[0] = tot[1] = 0;
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (mark[y]) continue;
    cnt[x] = 1, dis[x] = val[x];
    get_ans(y, x);
    cnt[x] = -1, dis[x] = 0;
    get_dis(y, x);
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (mark[y]) continue;
    solve(y);
  }
}
int main() {
  int a, b;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    ans = (ans + val[i]) % P;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    add_edge(a, b);
    add_edge(b, a);
  }
  solve(1);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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 M = 2e5 + 5, P = 1e9 + 7;
int n, val[M];
int head[M], asdf;
struct edge {
  int to, nxt;
} G[M * 2];
void add_edge(int a, int b) {
  G[++asdf].to = b;
  G[asdf].nxt = head[a];
  head[a] = asdf;
}
int Md(int x) { return (x % P + P) % P; }
bool mark[M];
int sz[M], son[M], que_len, que[M];
void get_root(int x, int f) {
  sz[x] = 1;
  son[x] = 0;
  que[++que_len] = x;
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_root(y, x);
    sz[x] += sz[y];
    if (sz[y] > son[x]) son[x] = sz[y];
  }
}
int ans, sum[2], tot[2];
int dis[M], cnt[M];
void get_ans(int x, int f) {
  cnt[x] = cnt[f] * (-1);
  dis[x] = Md(val[x] - dis[f]);
  if (cnt[x] == 1) {
    ans = Md(ans + 2ll * dis[x] % P);
    ans = Md(1ll * dis[x] * tot[1] % P + ans);
    ans = Md(-sum[1] + ans);
    ans = Md(1ll * dis[x] * tot[1] % P + ans);
    ans = Md(-sum[1] + ans);
  } else {
    ans = Md(1ll * dis[x] * tot[0] % P + ans);
    ans = Md(sum[0] + ans);
    ans = Md(1ll * dis[x] * tot[0] % P + ans);
    ans = Md(sum[0] + ans);
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_ans(y, x);
  }
}
void get_dis(int x, int f) {
  cnt[x] = cnt[f] * (-1);
  dis[x] = Md(dis[f] + val[x] * cnt[x]);
  if (cnt[x] == 1) {
    sum[0] = Md(sum[0] + dis[x]);
    tot[0]++;
    if (tot[0] >= P) tot[0] -= P;
  } else {
    sum[1] = Md(sum[1] + dis[x]);
    tot[1]++;
    if (tot[1] >= P) tot[1] -= P;
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (y == f || mark[y]) continue;
    get_dis(y, x);
  }
}
void solve(int x) {
  que_len = 0;
  get_root(x, x);
  for (int i = 1; i <= que_len; i++) {
    son[que[i]] = max(son[que[i]], que_len - sz[que[i]]);
    if (son[que[i]] < son[x]) x = que[i];
  }
  mark[x] = 1;
  sum[0] = sum[1] = 0;
  tot[0] = tot[1] = 0;
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (mark[y]) continue;
    cnt[x] = 1, dis[x] = val[x];
    get_ans(y, x);
    cnt[x] = -1, dis[x] = 0;
    get_dis(y, x);
  }
  for (int i = head[x]; i; i = G[i].nxt) {
    int y = G[i].to;
    if (mark[y]) continue;
    solve(y);
  }
}
int main() {
  int a, b;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    ans = (ans + val[i]) % P;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    add_edge(a, b);
    add_edge(b, a);
  }
  solve(1);
  printf("%d\n", ans);
  return 0;
}

```