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;
struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
struct ListNode {
  int val;
  ListNode* next;
  ListNode(int x) : val(x), next(NULL) {}
};
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long nxtLong() {
  long long x;
  cin >> x;
  return x;
}
string nxtStr() {
  string x;
  cin >> x;
  return x;
}
char nxtChar() {
  char x;
  cin >> x;
  return x;
}
double nxtDouble() {
  double x;
  cin >> x;
  return x;
}
long double nxtLDouble() {
  long double x;
  cin >> x;
  return x;
}
TreeNode* makeTree(int n) {
  queue<TreeNode*> tree_nodes;
  TreeNode* root;
  if (n) {
    root = new TreeNode(nxt());
    tree_nodes.push(root);
  } else {
    return NULL;
  }
  while (!tree_nodes.empty()) {
    TreeNode* node = tree_nodes.front();
    tree_nodes.pop();
    bool left;
    cin >> left;
    if (left) {
      node->left = new TreeNode(nxt());
      tree_nodes.push(node->left);
    }
    bool right;
    cin >> right;
    if (right) {
      node->right = new TreeNode(nxt());
      tree_nodes.push(node->right);
    }
  }
  return root;
}
ListNode* makeLinkedList(int n) {
  ListNode* list = new ListNode(0);
  ListNode* dummy = list;
  for (int i = 0; i < n; ++i) {
    list->next = new ListNode(nxt());
    list = list->next;
  }
  return dummy->next;
}
const int mod = 1000000007;
void solve() {
  int n = nxt();
  vector<int> parent(n + 1), sum(n + 1);
  for (int i = 1; i <= n; ++i) {
    parent[i] = i;
  }
  function<int(int)> get_par = [&](int x) {
    if (parent[x] == x) {
      return x;
    }
    int y = get_par(parent[x]);
    sum[x] += sum[parent[x]];
    sum[x] %= mod;
    parent[x] = y;
    return y;
  };
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    int k = nxt();
    for (int j = 0; j < k; ++j) {
      int v = nxt(), x = nxt();
      int u = get_par(v);
      parent[u] = i;
      sum[u] = x + sum[v];
      sum[u] %= mod;
      ans += sum[u];
      ans %= mod;
      if (ans <= 0) {
        ans += mod;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(9);
  solve();
  return 0;
}