#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int mo = 1e9 + 7;
const int maxn = 2e5 + 5;
int val[maxn];
int head[maxn], nxt[maxn << 1], to[maxn << 1];
int edgenum = -1;
void add(int u, int v) {
  edgenum++;
  nxt[edgenum] = head[u];
  head[u] = edgenum;
  to[edgenum] = v;
}
void add_edge(int u, int v) {
  add(u, v);
  add(v, u);
}
LL dp[maxn][2];
bool which[maxn];
LL ans = 0;
void dfs1(int now, int fa) {
  dp[now][0] = 1;
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    dfs1(to[i], now);
    dp[now][0] += dp[to[i]][1];
    dp[now][0] %= mo;
    dp[now][1] += dp[to[i]][0];
    dp[now][1] %= mo;
  }
  ans = (ans + 2 * dp[now][0] * val[now] % mo) % mo;
}
void dfs2(int now, int fa) {
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    which[to[i]] = which[now] ^ 1;
    dfs2(to[i], now);
  }
}
void calin(int now, int fa) {
  LL sum0, sum1;
  bool first = 1;
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    calin(to[i], now);
    if (first) {
      sum0 = dp[to[i]][0];
      sum1 = dp[to[i]][1];
      first = 0;
    } else {
      ans = (ans + 2 * sum0 * dp[to[i]][0] % mo * (-val[now])) % mo;
      sum0 += dp[to[i]][0];
      sum0 %= mo;
      ans = (ans + 2 * sum1 * dp[to[i]][1] % mo * val[now]) % mo;
      sum1 += dp[to[i]][1];
      sum1 %= mo;
    }
  }
}
void calout(int now, int fa) {
  if (which[now] == 0) {
    ans = (ans +
           2 * (dp[1][0] - dp[now][0]) * (dp[now][0]) % mo * (val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
    ans = (ans +
           2 * (dp[1][1] - dp[now][1]) * (dp[now][1]) % mo * (-val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
  } else {
    ans = (ans +
           2 * (dp[1][0] - dp[now][1]) * (dp[now][1]) % mo * (-val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
    ans = (ans +
           2 * (dp[1][1] - dp[now][0]) * (dp[now][0]) % mo * val[now] % mo) %
          mo;
    if (ans < 0) ans += mo;
  }
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    calout(to[i], now);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  memset(head, -1, sizeof(head));
  for (i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    ans -= val[i];
  }
  int u, v;
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  calin(1, 0);
  calout(1, 0);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
using LL = long long;
const int mo = 1e9 + 7;
const int maxn = 2e5 + 5;
int val[maxn];
int head[maxn], nxt[maxn << 1], to[maxn << 1];
int edgenum = -1;
void add(int u, int v) {
  edgenum++;
  nxt[edgenum] = head[u];
  head[u] = edgenum;
  to[edgenum] = v;
}
void add_edge(int u, int v) {
  add(u, v);
  add(v, u);
}
LL dp[maxn][2];
bool which[maxn];
LL ans = 0;
void dfs1(int now, int fa) {
  dp[now][0] = 1;
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    dfs1(to[i], now);
    dp[now][0] += dp[to[i]][1];
    dp[now][0] %= mo;
    dp[now][1] += dp[to[i]][0];
    dp[now][1] %= mo;
  }
  ans = (ans + 2 * dp[now][0] * val[now] % mo) % mo;
}
void dfs2(int now, int fa) {
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    which[to[i]] = which[now] ^ 1;
    dfs2(to[i], now);
  }
}
void calin(int now, int fa) {
  LL sum0, sum1;
  bool first = 1;
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    calin(to[i], now);
    if (first) {
      sum0 = dp[to[i]][0];
      sum1 = dp[to[i]][1];
      first = 0;
    } else {
      ans = (ans + 2 * sum0 * dp[to[i]][0] % mo * (-val[now])) % mo;
      sum0 += dp[to[i]][0];
      sum0 %= mo;
      ans = (ans + 2 * sum1 * dp[to[i]][1] % mo * val[now]) % mo;
      sum1 += dp[to[i]][1];
      sum1 %= mo;
    }
  }
}
void calout(int now, int fa) {
  if (which[now] == 0) {
    ans = (ans +
           2 * (dp[1][0] - dp[now][0]) * (dp[now][0]) % mo * (val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
    ans = (ans +
           2 * (dp[1][1] - dp[now][1]) * (dp[now][1]) % mo * (-val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
  } else {
    ans = (ans +
           2 * (dp[1][0] - dp[now][1]) * (dp[now][1]) % mo * (-val[now]) % mo) %
          mo;
    if (ans < 0) ans += mo;
    ans = (ans +
           2 * (dp[1][1] - dp[now][0]) * (dp[now][0]) % mo * val[now] % mo) %
          mo;
    if (ans < 0) ans += mo;
  }
  for (int i = head[now]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    calout(to[i], now);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  memset(head, -1, sizeof(head));
  for (i = 1; i <= n; i++) {
    scanf("%d", &val[i]);
    ans -= val[i];
  }
  int u, v;
  for (i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  calin(1, 0);
  calout(1, 0);
  printf("%lld\n", ans);
  return 0;
}

```