#include <bits/stdc++.h>
using namespace std;
int n, m, fir[200100], nex[200100 * 2], to[200100 * 2], fa[200100], siz[200100],
    que[200100], hd, tl, top, dep[200100];
long long f1[200100], f2[200100], f3[200100], f4[200100], ans, v[200100],
    t1[200100], t2[200100], t3[200100], t4[200100];
bool vis[200100];
void link(int x, int y) {
  to[++top] = y;
  nex[top] = fir[x];
  fir[x] = top;
}
int core(int x) {
  int y, i, p, mn = n + 13;
  for (hd = 0, fa[que[tl = 1] = x] = 0; hd ^ tl;)
    for (i = fir[x = que[++hd]]; i; i = nex[i])
      if (!vis[y = to[i]] && y != fa[x]) fa[que[++tl] = y] = x;
  for (hd = tl; hd; hd--) {
    x = que[hd];
    siz[x] = 1;
    int tmp = 0;
    for (i = fir[x]; i; i = nex[i])
      if (!vis[y = to[i]] && y != fa[x]) {
        siz[x] += siz[y];
        if (siz[y] > tmp) tmp = siz[y];
      }
    if (tl - siz[x] > tmp) tmp = tl - siz[x];
    if (tmp < mn) mn = tmp, p = x;
  }
  return p;
}
void dg(int x) {
  int y, i, j, c = core(x);
  long long s1 = 0, s2 = 0, s3 = 0, s4 = 0;
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) {
      t1[to[i]] = t2[to[i]] = t3[to[i]] = t4[to[i]] = 0;
      for (hd = 0, fa[que[tl = 1] = to[i]] = c, dep[to[i]] = 1; hd ^ tl;) {
        for (j = fir[x = que[++hd]]; j; j = nex[j])
          if (!vis[y = to[j]] && y != fa[x])
            fa[que[++tl] = y] = x, dep[y] = dep[x] + 1;
        t3[to[i]]++;
        t4[to[i]]++;
        if (dep[x] & 1)
          t2[to[i]]++;
        else
          t1[to[i]]++;
      }
      s1 += t1[to[i]];
      s2 += t2[to[i]];
      s3 += t3[to[i]];
      s4 += t4[to[i]];
    }
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) {
      long long r1 = 0, r2 = 0, r3 = 0, r4 = 0;
      f1[to[i]] = (v[c] - v[to[i]] + 1000000007) % 1000000007;
      f2[to[i]] = (v[to[i]] - v[c] + 1000000007) % 1000000007;
      f3[to[i]] = 1000000007 - v[to[i]];
      f4[to[i]] = v[to[i]];
      for (hd = 0, fa[que[tl = 1] = to[i]] = c; hd ^ tl;) {
        for (j = fir[x = que[++hd]]; j; j = nex[j])
          if (!vis[y = to[j]] && y != fa[x]) {
            fa[que[++tl] = y] = x;
            f1[y] = (f1[x] + (1 - (dep[y] & 1) * 2) * v[y] + 1000000007) %
                    1000000007;
            f2[y] = (f2[x] + ((dep[y] & 1) * 2 - 1) * v[y] + 1000000007) %
                    1000000007;
            f3[y] = (f3[x] + (1 - (dep[y] & 1) * 2) * v[y] + 1000000007) %
                    1000000007;
            f4[y] = (f4[x] + ((dep[y] & 1) * 2 - 1) * v[y] + 1000000007) %
                    1000000007;
          }
        r3 += f3[x];
        r4 += f4[x];
        if (dep[x] & 1)
          r2 += f2[x];
        else
          r1 += f1[x];
        r1 %= 1000000007;
        r2 %= 1000000007;
        r3 %= 1000000007;
        r4 %= 1000000007;
      }
      ans = (ans + r1 + r2 + r3 + t3[to[i]] * v[c]) % 1000000007;
      ans = (ans + r1 * (s3 - t3[to[i]]) % 1000000007 +
             r2 * (s4 - t4[to[i]]) % 1000000007 +
             r3 * (s1 - t1[to[i]]) % 1000000007 +
             r4 * (s2 - t2[to[i]]) % 1000000007) %
            1000000007;
    }
  vis[c] = 1;
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) dg(to[i]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%I64d", &v[i]), v[i] = (v[i] + 1000000007) % 1000000007,
                           ans = (ans + v[i]) % 1000000007;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    link(x, y);
    link(y, x);
  }
  dg(1);
  printf("%I64d", ans);
  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;
int n, m, fir[200100], nex[200100 * 2], to[200100 * 2], fa[200100], siz[200100],
    que[200100], hd, tl, top, dep[200100];
long long f1[200100], f2[200100], f3[200100], f4[200100], ans, v[200100],
    t1[200100], t2[200100], t3[200100], t4[200100];
bool vis[200100];
void link(int x, int y) {
  to[++top] = y;
  nex[top] = fir[x];
  fir[x] = top;
}
int core(int x) {
  int y, i, p, mn = n + 13;
  for (hd = 0, fa[que[tl = 1] = x] = 0; hd ^ tl;)
    for (i = fir[x = que[++hd]]; i; i = nex[i])
      if (!vis[y = to[i]] && y != fa[x]) fa[que[++tl] = y] = x;
  for (hd = tl; hd; hd--) {
    x = que[hd];
    siz[x] = 1;
    int tmp = 0;
    for (i = fir[x]; i; i = nex[i])
      if (!vis[y = to[i]] && y != fa[x]) {
        siz[x] += siz[y];
        if (siz[y] > tmp) tmp = siz[y];
      }
    if (tl - siz[x] > tmp) tmp = tl - siz[x];
    if (tmp < mn) mn = tmp, p = x;
  }
  return p;
}
void dg(int x) {
  int y, i, j, c = core(x);
  long long s1 = 0, s2 = 0, s3 = 0, s4 = 0;
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) {
      t1[to[i]] = t2[to[i]] = t3[to[i]] = t4[to[i]] = 0;
      for (hd = 0, fa[que[tl = 1] = to[i]] = c, dep[to[i]] = 1; hd ^ tl;) {
        for (j = fir[x = que[++hd]]; j; j = nex[j])
          if (!vis[y = to[j]] && y != fa[x])
            fa[que[++tl] = y] = x, dep[y] = dep[x] + 1;
        t3[to[i]]++;
        t4[to[i]]++;
        if (dep[x] & 1)
          t2[to[i]]++;
        else
          t1[to[i]]++;
      }
      s1 += t1[to[i]];
      s2 += t2[to[i]];
      s3 += t3[to[i]];
      s4 += t4[to[i]];
    }
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) {
      long long r1 = 0, r2 = 0, r3 = 0, r4 = 0;
      f1[to[i]] = (v[c] - v[to[i]] + 1000000007) % 1000000007;
      f2[to[i]] = (v[to[i]] - v[c] + 1000000007) % 1000000007;
      f3[to[i]] = 1000000007 - v[to[i]];
      f4[to[i]] = v[to[i]];
      for (hd = 0, fa[que[tl = 1] = to[i]] = c; hd ^ tl;) {
        for (j = fir[x = que[++hd]]; j; j = nex[j])
          if (!vis[y = to[j]] && y != fa[x]) {
            fa[que[++tl] = y] = x;
            f1[y] = (f1[x] + (1 - (dep[y] & 1) * 2) * v[y] + 1000000007) %
                    1000000007;
            f2[y] = (f2[x] + ((dep[y] & 1) * 2 - 1) * v[y] + 1000000007) %
                    1000000007;
            f3[y] = (f3[x] + (1 - (dep[y] & 1) * 2) * v[y] + 1000000007) %
                    1000000007;
            f4[y] = (f4[x] + ((dep[y] & 1) * 2 - 1) * v[y] + 1000000007) %
                    1000000007;
          }
        r3 += f3[x];
        r4 += f4[x];
        if (dep[x] & 1)
          r2 += f2[x];
        else
          r1 += f1[x];
        r1 %= 1000000007;
        r2 %= 1000000007;
        r3 %= 1000000007;
        r4 %= 1000000007;
      }
      ans = (ans + r1 + r2 + r3 + t3[to[i]] * v[c]) % 1000000007;
      ans = (ans + r1 * (s3 - t3[to[i]]) % 1000000007 +
             r2 * (s4 - t4[to[i]]) % 1000000007 +
             r3 * (s1 - t1[to[i]]) % 1000000007 +
             r4 * (s2 - t2[to[i]]) % 1000000007) %
            1000000007;
    }
  vis[c] = 1;
  for (i = fir[c]; i; i = nex[i])
    if (!vis[to[i]]) dg(to[i]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%I64d", &v[i]), v[i] = (v[i] + 1000000007) % 1000000007,
                           ans = (ans + v[i]) % 1000000007;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    link(x, y);
    link(y, x);
  }
  dg(1);
  printf("%I64d", ans);
  return 0;
}

```