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;
const long long mod = 1e9 + 7;
vector<long long> rt;
vector<long long> dep;
vector<pair<long long, long long> > wed;
long long depth(long long);
long long root(long long v) {
  if (rt[rt[v]] != v) {
    int prev = rt[v];
    rt[v] = root(prev);
    dep[v] = depth(prev) + dep[v];
    dep[v] = (dep[v] + mod) % mod;
  }
  return rt[v];
}
void hang(long long v, long long to) {
  rt[v] = to;
  dep[v] = wed[v].second;
}
long long depth(long long v) { return dep[v]; }
int main() {
  long long n;
  cin >> n;
  rt.resize(n);
  dep.resize(n);
  wed.resize(n);
  for (long long i = 0; i < n; i++) {
    rt[i] = i;
    dep[i] = 0;
  }
  for (long long i = 0; i < n; i++) {
    long long k;
    cin >> k;
    for (long long j = 0; j < k; j++) {
      long long v, x;
      cin >> v >> x;
      v--;
      long long rt = root(v);
      long long dep = depth(v);
      wed[rt] = pair<long long, long long>(i, (dep + x + mod) % mod);
      hang(rt, i);
    }
  }
  long long sum = 0;
  for (long long i = 0; i < n; i++) {
    sum += wed[i].second;
    sum = (sum + mod) % mod;
  }
  cout << sum;
  return 0;
}