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;
long long int n;
vector<pair<long long int, long long int> > in[100005];
vector<pair<long long int, pair<long long int, long long int> > > g[100005];
long long int par[100005];
long long int par_find(long long int v) {
  return par[v] == v ? par[v] : par[v] = par_find(par[v]);
}
long long int tin[100005], tout[100005], timer;
long long int mark[100005];
long long int ans, seg[1000001];
void update(long long int idx, long long int lo, long long int hi,
            long long int ps, long long int val) {
  if (lo == hi) {
    seg[idx] = val;
    return;
  }
  long long int mid = (lo + hi) / 2;
  if (ps <= mid)
    update(2 * idx, lo, mid, ps, val);
  else
    update(2 * idx + 1, mid + 1, hi, ps, val);
  seg[idx] = (seg[2 * idx] + seg[2 * idx + 1]) % 1000000007;
}
long long int query(long long int idx, long long int lo, long long int hi,
                    long long int st, long long int en) {
  if (lo == st && hi == en) return seg[idx];
  long long int mid = (lo + hi) / 2;
  if (en <= mid)
    return query(2 * idx, lo, mid, st, en);
  else if (st > mid)
    return query(2 * idx + 1, mid + 1, hi, st, en);
  else
    return (query(2 * idx, lo, mid, st, mid) +
            query(2 * idx + 1, mid + 1, hi, mid + 1, en)) %
           1000000007;
}
void dfs1(long long int st) {
  tin[st] = ++timer;
  mark[st] = 1;
  for (auto v : g[st]) dfs1(v.second.first);
  tout[st] = ++timer;
}
void dfs(long long int st) {
  mark[st] = 1;
  for (auto xp : g[st]) {
    dfs(xp.second.first);
    long long int qval =
        query(1, 1, 2 * n, tin[xp.second.first], tin[xp.second.second]);
    qval = (qval + xp.first) % 1000000007;
    ans = (ans + qval) % 1000000007;
    update(1, 1, 2 * n, tin[xp.second.first], qval);
    update(1, 1, 2 * n, tout[xp.second.first], 1000000007 - qval);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long int i = 1; i <= n; ++i) par[i] = i;
  for (long long int i = 1; i <= n; ++i) {
    long long int k;
    cin >> k;
    in[i].resize(k);
    for (long long int j = 0; j <= k - 1; ++j) {
      cin >> in[i][j].first >> in[i][j].second;
      in[i][j].second += 1000000007;
      in[i][j].second %= 1000000007;
      long long int pu = par_find(in[i][j].first);
      g[i].push_back(make_pair(in[i][j].second, make_pair(pu, in[i][j].first)));
      par[pu] = i;
    }
  }
  for (long long int i = n; i >= 1; i--)
    if (!mark[i]) dfs1(i);
  for (long long int i = 1; i <= n; ++i) mark[i] = 0;
  for (long long int i = n; i >= 1; i--)
    if (!mark[i]) dfs(i);
  cout << ans << '\n';
}