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, raod, p[100002], bst[100002], sz[100002], mod = 1000000007, x, y, z, t,
                                                 ad;
long long ans, kl[100002];
vector<pair<int, pair<int, int> > > g;
map<int, int> mp[100002];
vector<int> v[100002];
int find(int x) {
  if (p[x] == 0) return x;
  return p[x] = find(p[x]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    g.clear();
    scanf("%d", &raod);
    if (raod == 0) {
      bst[i] = i;
      v[i].push_back(i);
      sz[i] = 1;
      continue;
    }
    for (int j = 1; j <= raod; j++) {
      scanf("%d%d", &x, &y);
      z = find(x);
      g.push_back(make_pair(sz[z], make_pair(x, y)));
    }
    sort(g.begin(), g.end());
    reverse(g.begin(), g.end());
    int x = find(g[0].second.first);
    bst[i] = bst[x];
    v[bst[x]].push_back(i);
    ans +=
        (mp[bst[x]][g[0].second.first] + kl[bst[x]]) % mod + g[0].second.second;
    kl[bst[x]] +=
        mp[bst[x]][g[0].second.first] + kl[bst[x]] + g[0].second.second;
    ans %= mod;
    kl[bst[x]] %= mod;
    mp[bst[x]][i] = -kl[bst[x]];
    for (int j = 1; j < g.size(); j++) {
      t = bst[find(g[j].second.first)],
      ad =
          ((mp[t][g[j].second.first] + kl[t]) % mod + g[j].second.second) % mod;
      ans += ad;
      ans %= mod;
      for (int l = 0; l < v[t].size(); l++) {
        mp[bst[x]][v[t][l]] =
            ((mp[t][v[t][l]] + ad) % mod + kl[t]) % mod - kl[bst[x]];
        mp[bst[x]][v[t][l]] %= mod;
        v[bst[x]].push_back(v[t][l]);
      }
      v[t].clear();
    }
    for (int j = 0; j < g.size(); j++) {
      sz[i] += sz[find(g[j].second.first)];
      p[find(g[j].second.first)] = i;
    }
  }
  cout << (ans + mod) % mod << '\n';
  return 0;
}