#include <bits/stdc++.h>
using namespace std;
template <typename A>
inline A Max(const A &x, const A &y) {
  return x > y ? x : y;
}
template <typename A>
inline A Min(const A &x, const A &y) {
  return x < y ? x : y;
}
template <typename A>
inline void Swap(A &x, A &y) {
  x ^= y, y ^= x, x ^= y;
}
template <typename A>
inline A Abs(const A &x) {
  return x > 0 ? x : -x;
}
template <typename A>
inline void read(A &x) {
  char c;
  A neg = 1;
  do {
    c = getchar();
  } while ((c < '0' || c > '9') && c != '-');
  if (c == '-') neg = -1, c = getchar();
  x = 0;
  do {
    x = x * 10 + c - 48;
    c = getchar();
  } while (c >= '0' && c <= '9');
  x *= neg;
}
template <typename A, typename B>
inline void read(A &a, B &b) {
  read(a), read(b);
}
template <typename A, typename B, typename C>
inline void read(A &a, B &b, C &c) {
  read(a), read(b), read(c);
}
template <typename A, typename B, typename C, typename D>
inline void read(A &a, B &b, C &c, D &d) {
  read(a), read(b), read(c), read(d);
}
template <typename A>
inline void put(const A &tmp) {
  A x = tmp;
  if (x == 0) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  char buf[30];
  int buf_size = 0;
  while (x) {
    buf[++buf_size] = x % 10 + 48;
    x /= 10;
  }
  for (int i = (buf_size); i >= (1); --i) putchar(buf[i]);
}
inline int mul_mod(int a, int b, int mo) {
  int ret;
  __asm__ __volatile__("\tmull %%ebx\n\tdivl %%ecx\n"
                       : "=d"(ret)
                       : "a"(a), "b"(b), "c"(mo));
  return ret;
}
const int N = 200005;
const long long P = 1000000007;
int la[N], en[N << 1], ne[N << 1], top;
inline void add(int x, int y) {
  ne[++top] = la[x];
  en[top] = y;
  la[x] = top;
}
int val[N], n, x, y;
namespace problem {
inline void beTakenPlace() {}
int root, tsiz;
long long ans;
int siz[N], b[N];
inline void getroot(int u, int p) {
  siz[u] = 1;
  int maxi = 0;
  for (int i = la[u]; i; i = ne[i])
    if (en[i] != p && !b[en[i]]) {
      getroot(en[i], u);
      siz[u] += siz[en[i]];
      if (siz[en[i]] > maxi) maxi = siz[en[i]];
    }
  maxi = Max(maxi, tsiz - siz[u]);
  if (maxi < siz[0]) siz[0] = maxi, root = u;
}
int dep[N], oddsiz[N];
long long ptb[N], ntb[N];
long long pbt[N], nbt[N];
inline void main_dfs(int u, int p) {
  siz[u] = 1;
  oddsiz[u] = dep[u] & 1;
  ptb[u] = ntb[u] = pbt[u] = nbt[u] = 0;
  for (int i = la[u]; i; i = ne[i])
    if (en[i] != p && !b[en[i]]) {
      dep[en[i]] = dep[u] + 1;
      main_dfs(en[i], u);
      siz[u] += siz[en[i]];
      oddsiz[u] += oddsiz[en[i]];
      (ptb[u] += ptb[en[i]]) %= P, (ntb[u] += ntb[en[i]]) %= P;
      (pbt[u] += pbt[en[i]]) %= P, (nbt[u] += nbt[en[i]]) %= P;
    }
  if (dep[u] & 1) {
    (ptb[u] -= 1LL * val[u] * siz[u]) %= P;
    (ntb[u] += 1LL * val[u] * siz[u]) %= P;
    (pbt[u] -= 1LL * val[u] * (siz[u] - oddsiz[u])) %= P;
    (nbt[u] += 1LL * val[u] * oddsiz[u]) %= P;
  } else {
    (ptb[u] += 1LL * val[u] * siz[u]) %= P;
    (ntb[u] -= 1LL * val[u] * siz[u]) %= P;
    (pbt[u] += 1LL * val[u] * (siz[u] - oddsiz[u])) %= P;
    (nbt[u] -= 1LL * val[u] * oddsiz[u]) %= P;
  }
}
inline void solve(int u) {
  dep[u] = 0;
  main_dfs(u, 0);
  int v;
  for (int i = la[u]; i; i = ne[i])
    if (!b[en[i]]) {
      v = en[i];
      ans +=
          pbt[v] * (siz[u] - siz[v]) +
          (ptb[u] - ptb[v] - 1LL * val[u] * siz[v]) % P * (siz[v] - oddsiz[v]);
      ans %= P;
      ans += nbt[v] * (siz[u] - siz[v]) +
             (ntb[u] - ntb[v] + 1LL * val[u] * siz[v]) % P * (oddsiz[v]);
      ans %= P;
    }
  ans += ptb[u];
  ans %= P;
  b[u] = 1;
  for (int i = la[u]; i; i = ne[i])
    if (!b[en[i]]) {
      siz[0] = tsiz = siz[en[i]];
      getroot(en[i], u);
      solve(root);
    }
}
inline long long main() {
  siz[0] = tsiz = n;
  getroot(1, 0);
  solve(root);
  if (ans < 0) ans += P;
  return ans;
}
}  // namespace problem
int main() {
  read(n);
  for (register int i = (1); i <= (n); ++i) read(val[i]);
  for (int i = (1); i < (n); ++i) {
    read(x, y);
    add(x, y);
    add(y, x);
  }
  printf("%d", problem::main());
  return 0;
}

### Prompt

Your challenge is to write 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 <typename A>
inline A Max(const A &x, const A &y) {
  return x > y ? x : y;
}
template <typename A>
inline A Min(const A &x, const A &y) {
  return x < y ? x : y;
}
template <typename A>
inline void Swap(A &x, A &y) {
  x ^= y, y ^= x, x ^= y;
}
template <typename A>
inline A Abs(const A &x) {
  return x > 0 ? x : -x;
}
template <typename A>
inline void read(A &x) {
  char c;
  A neg = 1;
  do {
    c = getchar();
  } while ((c < '0' || c > '9') && c != '-');
  if (c == '-') neg = -1, c = getchar();
  x = 0;
  do {
    x = x * 10 + c - 48;
    c = getchar();
  } while (c >= '0' && c <= '9');
  x *= neg;
}
template <typename A, typename B>
inline void read(A &a, B &b) {
  read(a), read(b);
}
template <typename A, typename B, typename C>
inline void read(A &a, B &b, C &c) {
  read(a), read(b), read(c);
}
template <typename A, typename B, typename C, typename D>
inline void read(A &a, B &b, C &c, D &d) {
  read(a), read(b), read(c), read(d);
}
template <typename A>
inline void put(const A &tmp) {
  A x = tmp;
  if (x == 0) {
    putchar('0');
    return;
  }
  if (x < 0) putchar('-'), x = -x;
  char buf[30];
  int buf_size = 0;
  while (x) {
    buf[++buf_size] = x % 10 + 48;
    x /= 10;
  }
  for (int i = (buf_size); i >= (1); --i) putchar(buf[i]);
}
inline int mul_mod(int a, int b, int mo) {
  int ret;
  __asm__ __volatile__("\tmull %%ebx\n\tdivl %%ecx\n"
                       : "=d"(ret)
                       : "a"(a), "b"(b), "c"(mo));
  return ret;
}
const int N = 200005;
const long long P = 1000000007;
int la[N], en[N << 1], ne[N << 1], top;
inline void add(int x, int y) {
  ne[++top] = la[x];
  en[top] = y;
  la[x] = top;
}
int val[N], n, x, y;
namespace problem {
inline void beTakenPlace() {}
int root, tsiz;
long long ans;
int siz[N], b[N];
inline void getroot(int u, int p) {
  siz[u] = 1;
  int maxi = 0;
  for (int i = la[u]; i; i = ne[i])
    if (en[i] != p && !b[en[i]]) {
      getroot(en[i], u);
      siz[u] += siz[en[i]];
      if (siz[en[i]] > maxi) maxi = siz[en[i]];
    }
  maxi = Max(maxi, tsiz - siz[u]);
  if (maxi < siz[0]) siz[0] = maxi, root = u;
}
int dep[N], oddsiz[N];
long long ptb[N], ntb[N];
long long pbt[N], nbt[N];
inline void main_dfs(int u, int p) {
  siz[u] = 1;
  oddsiz[u] = dep[u] & 1;
  ptb[u] = ntb[u] = pbt[u] = nbt[u] = 0;
  for (int i = la[u]; i; i = ne[i])
    if (en[i] != p && !b[en[i]]) {
      dep[en[i]] = dep[u] + 1;
      main_dfs(en[i], u);
      siz[u] += siz[en[i]];
      oddsiz[u] += oddsiz[en[i]];
      (ptb[u] += ptb[en[i]]) %= P, (ntb[u] += ntb[en[i]]) %= P;
      (pbt[u] += pbt[en[i]]) %= P, (nbt[u] += nbt[en[i]]) %= P;
    }
  if (dep[u] & 1) {
    (ptb[u] -= 1LL * val[u] * siz[u]) %= P;
    (ntb[u] += 1LL * val[u] * siz[u]) %= P;
    (pbt[u] -= 1LL * val[u] * (siz[u] - oddsiz[u])) %= P;
    (nbt[u] += 1LL * val[u] * oddsiz[u]) %= P;
  } else {
    (ptb[u] += 1LL * val[u] * siz[u]) %= P;
    (ntb[u] -= 1LL * val[u] * siz[u]) %= P;
    (pbt[u] += 1LL * val[u] * (siz[u] - oddsiz[u])) %= P;
    (nbt[u] -= 1LL * val[u] * oddsiz[u]) %= P;
  }
}
inline void solve(int u) {
  dep[u] = 0;
  main_dfs(u, 0);
  int v;
  for (int i = la[u]; i; i = ne[i])
    if (!b[en[i]]) {
      v = en[i];
      ans +=
          pbt[v] * (siz[u] - siz[v]) +
          (ptb[u] - ptb[v] - 1LL * val[u] * siz[v]) % P * (siz[v] - oddsiz[v]);
      ans %= P;
      ans += nbt[v] * (siz[u] - siz[v]) +
             (ntb[u] - ntb[v] + 1LL * val[u] * siz[v]) % P * (oddsiz[v]);
      ans %= P;
    }
  ans += ptb[u];
  ans %= P;
  b[u] = 1;
  for (int i = la[u]; i; i = ne[i])
    if (!b[en[i]]) {
      siz[0] = tsiz = siz[en[i]];
      getroot(en[i], u);
      solve(root);
    }
}
inline long long main() {
  siz[0] = tsiz = n;
  getroot(1, 0);
  solve(root);
  if (ans < 0) ans += P;
  return ans;
}
}  // namespace problem
int main() {
  read(n);
  for (register int i = (1); i <= (n); ++i) read(val[i]);
  for (int i = (1); i < (n); ++i) {
    read(x, y);
    add(x, y);
    add(y, x);
  }
  printf("%d", problem::main());
  return 0;
}

```