#include <bits/stdc++.h>
using namespace std;
const long long N = 200000 + 5;
const long long MOD = 1000000007;
long long add(long long a, long long b) {
  a += b;
  a %= MOD;
  a += MOD;
  a %= MOD;
  return a;
}
long long mul(long long a, long long b) { return a * (long long)b % MOD; }
long long n;
long long v[N];
vector<long long> g[N];
long long dp[2][2][N];
long long cnt[2][2][N];
bool viz[N];
long long ans = 0;
long long tot[2][2];
long long tcn[2][2];
long long gt(long long nod, long long sgn, long long cnt1, long long s1,
             long long cnt2, long long s2) {
  long long ans = mul(s2, cnt1);
  ans = add(ans, mul(s1, cnt2));
  if (sgn == 0) {
    ans = add(ans, -mul(cnt1, mul(cnt2, v[nod])));
  } else {
    ans = add(ans, mul(cnt1, mul(cnt2, v[nod])));
  }
  return ans;
}
long long sub[N];
void dfs(long long nod) {
  sub[nod] = v[nod];
  viz[nod] = 1;
  dp[0][0][nod] = -v[nod];
  dp[1][1][nod] = v[nod];
  cnt[0][0][nod] = 1;
  cnt[1][1][nod] = 1;
  long long dad = -1;
  for (auto &nou : g[nod]) {
    if (viz[nou]) {
      dad = nou;
      continue;
    }
    sub[nod] += sub[nou];
    dfs(nou);
    for (long long st = 0; st < 2; st++) {
      for (long long now = 0; now < 2; now++) {
        cnt[st][now][nod] = add(cnt[st][now][nod], cnt[st][1 - now][nou]);
        if (now == 0) {
          dp[st][now][nod] =
              add(dp[st][now][nod], -mul(v[nod], cnt[st][1 - now][nou]));
        } else {
          dp[st][now][nod] =
              add(dp[st][now][nod], mul(v[nod], cnt[st][1 - now][nou]));
        }
        dp[st][now][nod] = add(dp[st][now][nod], dp[st][1 - now][nou]);
        if (nod == 1 && st == 1 && now == 1) {
        }
      }
    }
  }
  for (long long i = 0; i < 2; i++)
    for (long long j = 0; j < 2; j++) tot[i][j] = tcn[i][j] = 0;
  for (auto &nou : g[nod]) {
    if (nou == dad) continue;
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], cnt[i][j][nou]);
      }
    }
  }
  long long acum = 0;
  for (auto &nou : g[nod]) {
    if (nou == dad) continue;
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], -dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], -cnt[i][j][nou]);
      }
    }
    for (long long now = 0; now < 2; now++) {
      long long cur1 = gt(nod, now, cnt[1][1 - now][nou], dp[1][1 - now][nou],
                          tcn[0][1 - now], tot[0][1 - now]);
      long long cur2 = gt(nod, now, cnt[1][1 - now][nou], dp[1][1 - now][nou],
                          tcn[1][1 - now], tot[1][1 - now]);
      acum = add(acum, cur1);
      acum = add(acum, cur2);
    }
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], cnt[i][j][nou]);
      }
    }
  }
  ans = add(ans, acum);
  long long fini = 0;
  fini = add(fini, dp[1][1][nod]);
  fini = add(fini, dp[1][1][nod]);
  fini = add(fini, -sub[nod]);
  ans = add(ans, fini);
  return;
  return;
  cout << nod << " : " << dp[1][1][nod] << " " << dp[1][0][nod] << "\n";
  cout << nod << " : " << dp[0][1][nod] << " " << dp[0][0][nod] << "\n\n\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  cout << ans << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long N = 200000 + 5;
const long long MOD = 1000000007;
long long add(long long a, long long b) {
  a += b;
  a %= MOD;
  a += MOD;
  a %= MOD;
  return a;
}
long long mul(long long a, long long b) { return a * (long long)b % MOD; }
long long n;
long long v[N];
vector<long long> g[N];
long long dp[2][2][N];
long long cnt[2][2][N];
bool viz[N];
long long ans = 0;
long long tot[2][2];
long long tcn[2][2];
long long gt(long long nod, long long sgn, long long cnt1, long long s1,
             long long cnt2, long long s2) {
  long long ans = mul(s2, cnt1);
  ans = add(ans, mul(s1, cnt2));
  if (sgn == 0) {
    ans = add(ans, -mul(cnt1, mul(cnt2, v[nod])));
  } else {
    ans = add(ans, mul(cnt1, mul(cnt2, v[nod])));
  }
  return ans;
}
long long sub[N];
void dfs(long long nod) {
  sub[nod] = v[nod];
  viz[nod] = 1;
  dp[0][0][nod] = -v[nod];
  dp[1][1][nod] = v[nod];
  cnt[0][0][nod] = 1;
  cnt[1][1][nod] = 1;
  long long dad = -1;
  for (auto &nou : g[nod]) {
    if (viz[nou]) {
      dad = nou;
      continue;
    }
    sub[nod] += sub[nou];
    dfs(nou);
    for (long long st = 0; st < 2; st++) {
      for (long long now = 0; now < 2; now++) {
        cnt[st][now][nod] = add(cnt[st][now][nod], cnt[st][1 - now][nou]);
        if (now == 0) {
          dp[st][now][nod] =
              add(dp[st][now][nod], -mul(v[nod], cnt[st][1 - now][nou]));
        } else {
          dp[st][now][nod] =
              add(dp[st][now][nod], mul(v[nod], cnt[st][1 - now][nou]));
        }
        dp[st][now][nod] = add(dp[st][now][nod], dp[st][1 - now][nou]);
        if (nod == 1 && st == 1 && now == 1) {
        }
      }
    }
  }
  for (long long i = 0; i < 2; i++)
    for (long long j = 0; j < 2; j++) tot[i][j] = tcn[i][j] = 0;
  for (auto &nou : g[nod]) {
    if (nou == dad) continue;
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], cnt[i][j][nou]);
      }
    }
  }
  long long acum = 0;
  for (auto &nou : g[nod]) {
    if (nou == dad) continue;
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], -dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], -cnt[i][j][nou]);
      }
    }
    for (long long now = 0; now < 2; now++) {
      long long cur1 = gt(nod, now, cnt[1][1 - now][nou], dp[1][1 - now][nou],
                          tcn[0][1 - now], tot[0][1 - now]);
      long long cur2 = gt(nod, now, cnt[1][1 - now][nou], dp[1][1 - now][nou],
                          tcn[1][1 - now], tot[1][1 - now]);
      acum = add(acum, cur1);
      acum = add(acum, cur2);
    }
    for (long long i = 0; i < 2; i++) {
      for (long long j = 0; j < 2; j++) {
        tot[i][j] = add(tot[i][j], dp[i][j][nou]);
        tcn[i][j] = add(tcn[i][j], cnt[i][j][nou]);
      }
    }
  }
  ans = add(ans, acum);
  long long fini = 0;
  fini = add(fini, dp[1][1][nod]);
  fini = add(fini, dp[1][1][nod]);
  fini = add(fini, -sub[nod]);
  ans = add(ans, fini);
  return;
  return;
  cout << nod << " : " << dp[1][1][nod] << " " << dp[1][0][nod] << "\n";
  cout << nod << " : " << dp[0][1][nod] << " " << dp[0][0][nod] << "\n\n\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  cout << ans << "\n";
  return 0;
}

```