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;
int n;
pair<int, int> prv[100007];
long long sm[100007];
int br[100007];
bool used[100007];
pair<int, int> find(pair<int, int> x) {
  if (prv[x.first].first == -1) {
    return x;
  } else {
    pair<int, int> ret = find(make_pair(
        prv[x.first].first, (prv[x.first].second + x.second) % 1000000007));
    prv[x.first].first = ret.first;
    prv[x.first].second = (ret.second - x.second + 1000000007) % 1000000007;
    return ret;
  }
}
void input() {
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; i++) {
    prv[i].first = -1;
    prv[i].second = 0;
    br[i] = 1;
    sm[i] = 0;
  }
  for (i = 1; i <= n; i++) {
    int sz;
    int x, y;
    scanf("%d", &sz);
    while (sz != 0) {
      sz--;
      scanf("%d%d", &x, &y);
      if (y < 0) {
        y += 1000000007;
      }
      pair<int, int> p;
      p = find(make_pair(x, 0));
      sm[i] += sm[p.first];
      sm[i] %= 1000000007;
      br[i] += br[p.first];
      sm[i] += p.second + y;
      sm[i] %= 1000000007;
      prv[p.first].first = i;
      prv[p.first].second = (p.second + y) % 1000000007;
    }
  }
}
void solve() {
  int i;
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    int k = find(make_pair(i, 0)).first;
    if (used[k] == false) {
      ans = (ans + sm[k]) % 1000000007;
    }
    used[k] = true;
  }
  printf("%d\n", ans);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}