Problem: An oriented weighted forest is an acyclic weighted digraph in which from each vertex at most one edge goes.

The root of vertex v of an oriented weighted forest is a vertex from which no edge goes and which can be reached from vertex v moving along the edges of the weighted oriented forest. We denote the root of vertex v as root(v).

The depth of vertex v is the sum of weights of paths passing from the vertex v to its root. Let's denote the depth of the vertex v as depth(v).

Let's consider the process of constructing a weighted directed forest. Initially, the forest does not contain vertices. Vertices are added sequentially one by one. Overall, there are n performed operations of adding. The i-th (i > 0) adding operation is described by a set of numbers (k, v1, x1, v2, x2, ... , vk, xk) and means that we should add vertex number i and k edges to the graph: an edge from vertex root(v1) to vertex i with weight depth(v1) + x1, an edge from vertex root(v2) to vertex i with weight depth(v2) + x2 and so on. If k = 0, then only vertex i is added to the graph, there are no added edges.

Your task is like this: given the operations of adding vertices, calculate the sum of the weights of all edges of the forest, resulting after the application of all defined operations, modulo 1000000007 (109 + 7).

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of operations of adding a vertex.

Next n lines contain descriptions of the operations, the i-th line contains the description of the operation of adding the i-th vertex in the following format: the first number of a line is an integer k (0 ≤ k ≤ i - 1), then follow 2k space-separated integers: v1, x1, v2, x2, ... , vk, xk (1 ≤ vj ≤ i - 1, |xj| ≤ 109). 

The operations are given in the order, in which they should be applied to the graph. It is guaranteed that sum k of all operations does not exceed 105, also that applying operations of adding vertexes does not result in loops and multiple edges. 

Output

Print a single number — the sum of weights of all edges of the resulting graph modulo 1000000007 (109 + 7).

Examples

Input

6
0
0
1 2 1
2 1 5 2 2
1 1 2
1 3 4


Output

30


Input

5
0
1 1 5
0
0
2 3 1 4 3


Output

9

Note

Conside the first sample:

  1. Vertex 1 is added. k = 0, thus no edges are added.
  2. Vertex 2 is added. k = 0, thus no edges are added.
  3. Vertex 3 is added. k = 1. v1 = 2, x1 = 1. Edge from vertex root(2) = 2 to vertex 3 with weight depth(2) + x1 = 0 + 1 = 1 is added. 
  4. Vertex 4 is added. k = 2. v1 = 1, x1 = 5. Edge from vertex root(1) = 1 to vertex 4 with weight depth(1) + x1 = 0 + 5 = 5 is added. v2 = 2, x2 = 2. Edge from vertex root(2) = 3 to vertex 4 with weight depth(2) + x1 = 1 + 2 = 3 is added.
  5. Vertex 5 is added. k = 1. v1 = 1, x1 = 2. Edge from vertex root(1) = 4 to vertex 5 with weight depth(1) + x1 = 5 + 2 = 7 is added.
  6. Vertex 6 is added. k = 1. v1 = 3, x1 = 4. Edge from vertex root(3) = 5 to vertex 6 with weight depth(3) + x1 = 10 + 4 = 14 is added.



The resulting graph is shown on the pictore below: <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
void splitstr(const string &s, vector<T> &out) {
  istringstream in(s);
  out.clear();
  copy(istream_iterator<T>(in), istream_iterator<T>(), back_inserter(out));
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
static void redirect(int argc, const char **argv) {
  if (argc > 1) {
    static filebuf f;
    f.open(argv[1], ios::in);
    cin.rdbuf(&f);
    if (!cin) {
      cerr << "Failed to open '" << argv[1] << "'" << endl;
      exit(1);
    }
  }
  if (argc > 2) {
    static filebuf f;
    f.open(argv[2], ios::out | ios::trunc);
    cout.rdbuf(&f);
    if (!cout) {
      cerr << "Failed to open '" << argv[2] << "'" << endl;
    }
  }
}
static pair<int, long long> find(vector<int> &parent, vector<long long> &weight,
                                 int x) {
  int y = x;
  long long w = 0;
  while (parent[y] >= 0) {
    w = (w + weight[y]) % 1000000007;
    y = parent[y];
  }
  long long ws = 0;
  while (parent[x] >= 0) {
    int n = parent[x];
    long long wn = weight[x];
    parent[x] = y;
    weight[x] = (w - ws) % 1000000007;
    ws = (ws + wn) % 1000000007;
    x = n;
  }
  return make_pair(y, w);
}
int main(int argc, const char **argv) {
  redirect(argc, argv);
  int N;
  cin >> N;
  vector<int> parent(N, -1);
  vector<long long> weight(N, 0);
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    int K;
    cin >> K;
    for (int j = 0; j < K; j++) {
      int v;
      long long x;
      cin >> v >> x;
      v--;
      pair<int, long long> root = find(parent, weight, v);
      long long w = (root.second + x) % 1000000007;
      ans = (ans + w) % 1000000007;
      parent[root.first] = i;
      weight[root.first] = w;
      cerr << i << ' ' << root.first << ' ' << w << '\n';
    }
  }
  ans %= 1000000007;
  if (ans < 0) ans += 1000000007;
  cout << ans << '\n';
  return 0;
}