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;
vector<pair<pair<int, int>, int> > quer[100010];
int par[100010], sz[100010], in[100010], val[100010],
    ct = 0, tree[4 * 100010], lazy[4 * 100010], visited[100010];
vector<int> graph[100010];
pair<int, int> rng[100010];
int mult(int a, int b) { return 1LL * a * b % 1000000007; }
int add(int a, int b) { return (a + b) % 1000000007; }
int root(int a) {
  while (par[a] != a) {
    a = par[a];
    par[a] = par[par[a]];
  }
  return a;
}
void unionit(int a, int b) {
  int x = root(a);
  int y = root(b);
  if (sz[x] < sz[y]) {
    par[x] = y;
    sz[y] += sz[x];
  } else {
    par[y] = x;
    sz[x] += sz[y];
  }
}
void dfs(int s, int par) {
  visited[s] = 1;
  in[s] = ct++;
  for (auto to : graph[s]) {
    if (to != par) {
      dfs(to, s);
    }
  }
  rng[s] = make_pair(in[s], ct - 1);
}
void push(int low, int high, int node) {
  if (lazy[node]) {
    tree[node] = add(tree[node], mult(high - low + 1, lazy[node]));
    if (low != high) {
      lazy[2 * node] = add(lazy[2 * node], lazy[node]);
      lazy[2 * node + 1] = add(lazy[2 * node + 1], lazy[node]);
    }
    lazy[node] = 0;
  }
}
void update(int low, int high, int begin, int end, int val, int node) {
  push(low, high, node);
  if (low > end || high < begin) return;
  if (low >= begin & high <= end) {
    lazy[node] = val;
    push(low, high, node);
    return;
  }
  int mid = (low + high) / 2;
  update(low, mid, begin, end, val, 2 * node);
  update(mid + 1, high, begin, end, val, 2 * node + 1);
  tree[node] = add(tree[2 * node], tree[2 * node + 1]);
}
int query(int low, int high, int begin, int end, int node) {
  push(low, high, node);
  if (low > end || high < begin) return 0;
  if (low >= begin && high <= end) return tree[node];
  int mid = (low + high) / 2;
  return add(query(low, mid, begin, end, 2 * node),
             query(mid + 1, high, begin, end, 2 * node + 1));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) par[i] = i, sz[i] = 1;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    for (int j = 0; j < k; j++) {
      int v, x;
      cin >> v >> x;
      x = (x + 1000000007) % 1000000007;
      v--;
      int y = val[root(v)];
      unionit(v, i);
      graph[y].push_back(i);
      graph[i].push_back(y);
      quer[i].push_back(make_pair(make_pair(v, x), y));
    }
    val[root(i)] = i;
  }
  for (int i = n - 1; i >= 0; i--) {
    if (!visited[i]) {
      dfs(i, -1);
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < quer[i].size(); j++) {
      int y = query(0, n - 1, in[quer[i][j].first.first],
                    in[quer[i][j].first.first], 1);
      ans = add(ans, add(y, quer[i][j].first.second));
      update(0, n - 1, rng[quer[i][j].second].first,
             rng[quer[i][j].second].second, add(y, quer[i][j].first.second), 1);
    }
  }
  cout << ans << '\n';
}