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 p[100100];
int sumtree[100100];
int sum[100100];
int find(int x) {
  if (x == p[x]) return x;
  return p[x] = find(p[x]);
}
void join(int x, int y) {
  x = find(x);
  y = find(y);
  if (rand() & 1)
    p[x] = y;
  else
    p[y] = x;
}
vector<vector<pair<int, int> > > g;
int mod = 1000000007;
int num[100100];
map<int, int> mm[100100];
int add[100100];
int main() {
  int n;
  scanf("%d", &n);
  g.resize(n);
  int res = 0;
  int last = 0;
  for (int i = (0); i < (n); ++i) {
    int k;
    scanf("%d", &k);
    if (k == 0) {
      mm[last][i] = 0;
      num[i] = last;
      last++;
      continue;
    }
    for (int j = (0); j < (k); ++j) {
      int v, x;
      scanf("%d%d", &v, &x);
      while (x < 0) x += mod;
      g[i].push_back(make_pair(v - 1, x));
    }
    int max1 = -1, v = -1, ver = -1;
    for (int j = (0); j < (k); ++j) {
      int c = g[i][j].first;
      if ((int)mm[num[c]].size() > max1) {
        max1 = mm[num[c]].size();
        v = num[c];
        ver = j;
      }
    }
    swap(g[i][ver], g[i][0]);
    int val = g[i][0].second;
    val += mm[v][g[i][0].first];
    if (val >= mod) val -= mod;
    val += add[v];
    if (val >= mod) val -= mod;
    res += val;
    if (res >= mod) res -= mod;
    add[v] += val;
    if (add[v] >= mod) add[v] -= mod;
    for (int j = (1); j < (k); ++j) {
      int c = num[g[i][j].first];
      int val = mm[c][g[i][j].first] + add[c];
      if (val >= mod) val -= mod;
      res += val;
      int x = g[i][j].second;
      if (res >= mod) res -= mod;
      res += x;
      if (res >= mod) res -= mod;
      for (map<int, int>::iterator it = mm[c].begin(); it != mm[c].end();
           ++it) {
        int nv = it->first;
        int nval = it->second + val;
        if (nval >= mod) nval -= mod;
        nval += x;
        if (nval >= mod) nval -= mod;
        nval += add[c];
        if (nval >= mod) nval -= mod;
        nval -= add[v];
        if (nval < 0) nval += mod;
        mm[v][nv] = nval;
        num[nv] = v;
      }
    }
    int q = -add[v];
    while (q < 0) q += mod;
    mm[v][i] = q;
    num[i] = v;
  }
  printf("%d\n", res);
  return 0;
}