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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
bool rev[200200];
int pa[200200], son[200200][2], val[200200], sum[200200];
void init_LCT(int n) {
  for (int i = 0; i <= n; i++) {
    rev[i] = val[i] = sum[i] = 0;
    pa[i] = son[i][0] = son[i][1] = 0;
  }
}
void push_up(int first) {
  int L = son[first][0], R = son[first][1];
  sum[first] = ((long long)val[first] + sum[L] + sum[R]) % 1000000007;
  if (sum[first] < 0) sum[first] += 1000000007;
}
void push_down(int first) {
  if (rev[first]) {
    swap(son[first][0], son[first][1]);
    rev[son[first][0]] ^= 1;
    rev[son[first][1]] ^= 1;
    rev[first] = 0;
  }
}
bool isroot(int first) {
  return son[pa[first]][0] != first && son[pa[first]][1] != first;
}
void down(int first) {
  if (!isroot(first)) down(pa[first]);
  push_down(first);
}
void rot(int first, int d) {
  int second = pa[first], z = pa[second];
  son[second][!d] = son[first][d];
  pa[son[first][d]] = second;
  pa[first] = pa[second];
  if (son[z][0] == second)
    son[z][0] = first;
  else if (son[z][1] == second)
    son[z][1] = first;
  pa[second] = first;
  son[first][d] = second;
  push_up(second);
}
void splay(int first) {
  down(first);
  while (!isroot(first)) {
    if (isroot(pa[first]))
      rot(first, son[pa[first]][0] == first);
    else {
      int second = pa[first], z = pa[second];
      int d = (son[z][0] == second);
      son[second][d] == first ? rot(first, !d) : rot(second, d);
      rot(first, d);
    }
  }
  push_up(first);
}
void access(int first) {
  for (int second = 0; first; first = pa[first]) {
    splay(first);
    son[first][1] = second;
    push_up(second = first);
  }
}
void make_root(int first) {
  access(first);
  splay(first);
  rev[first] ^= 1;
}
void link(int first, int second) {
  make_root(first);
  splay(first);
  pa[first] = second;
}
void cut(int first, int second) {
  make_root(first);
  splay(second);
  pa[son[second][0]] = pa[second];
  pa[second] = son[second][0] = 0;
}
int getpa(int v) {
  access(v);
  splay(v);
  while (son[v][0]) v = son[v][0];
  splay(v);
  return v;
}
int main() {
  int n;
  cin >> n;
  int cnt = n;
  int k, v, first, w;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &k);
    while (k--) {
      scanf("%d %d", &v, &first);
      access(v);
      splay(v);
      w = (first + sum[v]) % 1000000007;
      if (w < 0) w += 1000000007;
      ans += w;
      if (ans >= 1000000007) ans -= 1000000007;
      ++cnt;
      val[cnt] = w;
      pa[v] = cnt;
      pa[cnt] = i;
    }
  }
  printf("%d\n", ans);
  return 0;
}