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 double PI = 3.1415926535897;
const long long INF_LL = 1e18 + 7;
const double EPS = 1e-5;
const int INF_int = 1e9 + 7;
const int maxn = 1e5 + 10;
const int mod2 = 1e9 + 9;
const int mod3 = 1e9 + 33;
int p;
const int K = 47500;
pair<int, int> par[maxn];
pair<int, int> getpar(int x) {
  if (x == par[x].first) return par[x];
  par[x].second = (par[x].second + getpar(par[x].first).second) % INF_int;
  par[x].first = getpar(par[x].first).first;
  return par[x];
}
void unite(int v, int y, int w) {
  int x = getpar(v).first;
  int wei = getpar(v).second;
  wei += w;
  if (wei < 0) wei += INF_int;
  par[x] = {y, wei % INF_int};
  return;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < maxn; ++i) par[i].first = i;
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int k;
    cin >> k;
    for (int j = 0; j < k; ++j) {
      int v, x;
      cin >> v >> x;
      int t = getpar(v).first;
      unite(v, i + 1, x);
      ans = (ans + getpar(t).second) % INF_int;
    }
  }
  cout << ans;
}
int32_t main() {
  cout << fixed << setprecision(15);
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}