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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Solution implements Runnable {
    static int[] parent;
    static long[] w;
    static long MOD = 1000000007;
    
    static void update(int v) {
        if (parent[v] != -1) {
            update(parent[v]);
            w[v] = (w[v] + w[parent[v]]) % MOD;
            if (parent[parent[v]] != -1) parent[v] = parent[parent[v]];
        }
    }
    
    @Override
    public void run() {
        FastScanner in = new FastScanner(System.in);
        int n = in.nextInt();
        parent = new int[n];
        Arrays.fill(parent, -1);
        w = new long[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            int k = in.nextInt();
            for (int j = 0; j < k; j++) {
                int v = in.nextInt() - 1;
                long x = (in.nextInt() + MOD) % MOD;
                update(v);
                int r = parent[v];
                if (r == -1) r = v;
                parent[r] = i;
                sum = (sum + w[v] + x + 2 * MOD) % MOD;
                w[r] = (w[v] + x + MOD) % MOD;
            }
        }
        System.out.println(sum);
    }
    
    public static void main(String[] args) {
        new Thread(null, new Solution(), "1", 10000000).start();
    }
}

final class FastScanner {
    private StringTokenizer stringTokenizer = null;
    private BufferedReader bufferedReader; 
    
    public FastScanner(InputStream in) {
        bufferedReader = new BufferedReader(new InputStreamReader(in));
        try {
            stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    
    public String next() {
        while (!stringTokenizer.hasMoreTokens())
            try {
                stringTokenizer = new StringTokenizer(bufferedReader.readLine());
            } catch (IOException e) {
                e.printStackTrace();
                return null;
            }
        return stringTokenizer.nextToken();
    }
    
    public int nextInt() {
        return Integer.parseInt(next());
    }
    
    public long nextLong() {
        return Long.parseLong(next());
    }
    
    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public boolean hasNext() {
        return stringTokenizer.hasMoreTokens();
    }
    
    public void nextLine() {
        try {
            stringTokenizer = new StringTokenizer(bufferedReader.readLine());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}