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 int maxn = 1e5 + 9;
const long long mod = 1e9 + 7;
vector<pair<int, int> > vec;
int num[maxn], par[maxn];
int root(int v) { return ((par[v] < 0) ? v : root(par[v])); }
int sum(int v) { return (num[v] + ((par[v] < 0) ? 0 : sum(par[v]))) % mod; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(par, -1, sizeof par);
  int n, res = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    if (k == 0) continue;
    int mx, mxsz = 0, mxval;
    vec.clear();
    for (int t = 0; t < k; t++) {
      int v, x;
      cin >> v >> x;
      --v;
      vec.push_back(make_pair(root(v), (x + sum(v)) % mod));
      res = (res + vec.back().second) % mod;
      if (-par[vec.back().first] > mxsz) {
        mxsz = -par[vec.back().first];
        mx = vec.back().first;
        mxval = vec.back().second;
      }
    }
    num[mx] = (num[mx] + mxval) % mod;
    par[mx] -= 1;
    num[i] -= num[mx];
    par[i] = mx;
    for (auto x : vec) {
      int u = x.first, val = x.second;
      if (u == mx) continue;
      par[mx] += par[u];
      par[u] = mx;
      num[u] = ((num[u] + val) % mod - num[mx]) % mod;
    }
  }
  cout << (res + mod) % mod << "\n";
}