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 N = 110000;
vector<int> v[N];
int parent[N];
int in[N];
bool vis[N];
int order[2 * N];
int ind[N];
pair<int, int> times[N];
vector<pair<int, int> > e[N];
int t[4 * N];
int add[4 * N];
int sum[N];
int MOD = 1000000007;
int tm = 0;
int find(int k) {
  if (parent[k] == k)
    return parent[k];
  else
    parent[k] = find(parent[k]);
  return parent[k];
}
void dfs(int k) {
  times[k].first = tm;
  order[tm] = k;
  ind[k] = tm;
  tm++;
  vis[k] = true;
  for (int i = 0; i < v[k].size(); i++)
    if (!vis[v[k][i]]) dfs(v[k][i]);
  times[k].second = tm - 1;
}
void push(int v, int tl, int tr) {
  if (tl == tr) {
    t[v] = ((t[v] + add[v]) % MOD + MOD) % MOD;
    add[v] = 0;
  } else {
    int m = (tl + tr) / 2;
    t[v] += ((1ll * add[v] * (tr - tl + 1)) % MOD + MOD) % MOD;
    add[v * 2] = ((add[v * 2] + add[v]) % MOD + MOD) % MOD;
    add[v * 2 + 1] = ((add[v * 2 + 1] + add[v]) % MOD + MOD) % MOD;
    add[v] = 0;
  }
}
void update(int v, int tl, int tr, int l, int r, int val) {
  push(v, tl, tr);
  if (l > r) return;
  if (tl == l && tr == r) {
    add[v] = ((add[v] + val) % MOD + MOD) % MOD;
    push(v, tl, tr);
  } else {
    int m = (tl + tr) / 2;
    update(v * 2, tl, m, l, min(m, r), val);
    update(v * 2 + 1, m + 1, tr, max(l, m + 1), r, val);
    t[v] = ((t[v * 2] + t[v * 2 + 1]) % MOD + MOD) % MOD;
  }
}
int get(int v, int tl, int tr, int l, int r) {
  push(v, tl, tr);
  if (l > r) return 0;
  if (tl == l && tr == r) return t[v];
  int m = (tl + tr) / 2;
  return ((get(v * 2, tl, m, l, min(m, r)) +
           get(v * 2 + 1, m + 1, tr, max(m + 1, l), r)) %
              MOD +
          MOD) %
         MOD;
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    parent[i] = i;
    in[i] = 0;
    vis[i] = false;
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    for (int j = 0; j < k; j++) {
      int x1, x2;
      cin >> x1 >> x2;
      x1--;
      int to = find(x1);
      v[i].push_back(to);
      parent[to] = i;
      in[to]++;
      e[i].push_back(make_pair(x1, x2));
      cnt++;
    }
  }
  for (int i = 0; i < n; i++)
    if (in[i] == 0) dfs(i);
  for (int i = 0; i < 4 * n; i++) {
    t[i] = 0;
    sum[i] = 0;
    add[i] = 0;
  }
  for (int i = 0; i < n; i++) parent[i] = i;
  int answer = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < e[i].size(); j++) {
      int to = find(e[i][j].first);
      parent[to] = i;
      int ans = ((get(1, 0, n - 1, ind[e[i][j].first], ind[e[i][j].first]) +
                  e[i][j].second) %
                     MOD +
                 MOD) %
                MOD;
      update(1, 0, n - 1, times[to].first, times[to].second, ans);
      answer = ((ans + answer) % MOD + MOD) % MOD;
    }
  }
  cout << answer << "\n";
  return 0;
}