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;
const long long maxn = 5e5 + 100;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
long long n, k, dsu[maxn], t, st[maxn], ft[maxn], seg[maxn], ans;
vector<pair<long long, long long> > e[maxn];
vector<long long> g[maxn];
void init() {
  for (long long i = 1; i <= n; i++) {
    dsu[i] = i;
  }
}
long long par(long long v) {
  if (dsu[v] != v) dsu[v] = par(dsu[v]);
  return dsu[v];
}
void dfs(long long v) {
  st[v] = t++;
  for (long long i = 0; i < g[v].size(); i++) {
    dfs(g[v][i]);
  }
  ft[v] = t;
}
void upd(long long id, long long s, long long e, long long l, long long r,
         long long x) {
  if (r <= s || e <= l) return;
  if (l <= s && e <= r) {
    seg[id] += x;
    return;
  }
  long long mid = (s + e) / 2;
  upd(id * 2, s, mid, l, r, x);
  upd(id * 2 + 1, mid, e, l, r, x);
  return;
}
long long get(long long id, long long s, long long e, long long l,
              long long r) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r) return seg[id];
  long long mid = (s + e) / 2;
  return get(id * 2, s, mid, l, r) + get(id * 2 + 1, mid, e, l, r) + seg[id];
}
int main() {
  long long v, u, x;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  init();
  for (long long i = 1; i <= n; i++) {
    cin >> k;
    for (long long j = 0; j < k; j++) {
      cin >> v >> u;
      e[i].push_back({v, u});
      g[i].push_back(par(v));
      dsu[par(v)] = i;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (dsu[i] == i) dfs(i);
  }
  init();
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < e[i].size(); j++) {
      v = e[i][j].first;
      u = e[i][j].second;
      x = get(1, 0, n, st[v], st[v] + 1);
      v = par(v);
      upd(1, 0, n, st[v], ft[v], (x + u) % mod);
      ans += (x + u) % mod;
      ans %= mod;
      dsu[v] = i;
    }
  }
  cout << (ans + mod) % mod;
  return 0;
}