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;
inline void rinput(int a[], int n) {
  for (long i = 0; i < n; i++) scanf("%d", a + i);
}
const int INF = int(1e9);
const double EPS = 1e-5;
const long long INF64 = (long long)(1e18) + 1, md = 1000000007ll;
long long n, m, p[500005], k, x, v, g, l;
long long w[500005], nxt[500005], ans = 0;
void add(long long x) {
  p[x] = x;
  nxt[x] = w[x] = 0;
}
long long getroot(long long v) {
  if (p[v] == v) return v;
  long long t = getroot(p[v]);
  w[v] += w[p[v]];
  w[p[v]] %= md;
  nxt[v] %= md;
  w[v] %= md;
  p[v] = t;
  return p[v];
}
int main() {
  scanf("%I64d", &n);
  for (long i = 0; i < n; i++) {
    scanf("%I64d", &k);
    add(i);
    for (long j = 0; j < k; j++) {
      scanf("%I64d %I64d", &v, &x);
      v--;
      g = getroot(v);
      w[v] %= md;
      nxt[g] = x + w[v];
      nxt[g] %= md;
      p[g] = i;
      w[g] = w[v] + x;
    }
    nxt[i] %= md;
  }
  for (long i = 0; i < n; i++) {
    nxt[i] %= md;
    ans += nxt[i];
    ans %= md;
  }
  while (ans < 0) ans += md;
  ans %= md;
  printf("%I64d", ans);
  return 0;
}