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;
using namespace std;
inline long long __gcd(long long a, long long b);
void pdash(long long n = 1);
long long bitcount(long long u);
long long power(long long x, long long y);
long long power(long long x, long long y, long long z);
long long modInverse(long long n, long long p);
long long nCrF(long long n, long long r, long long p);
vector<long long> cordinate_compression(vector<long long>& v);
void make_uniquee(vector<long long>& vec);
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
long long myrand(long long mod = 1000000009);
void custome_random_shuffle(vector<long long>& vec) {
  shuffle(vec.begin(), vec.end(), rng);
  return;
}
void svec(vector<long long>& vec);
void smap(vector<long long, long long>& mp);
void sset(set<long long>& st);
const long long MAXN = 100005;
const long long MOD = 1000000007;
long long N, mom[MAXN], V[MAXN], ans;
long long find(long long n) {
  if (mom[n] == n) return n;
  long long k = mom[n];
  mom[n] = find(mom[n]);
  V[n] = (V[n] + V[k]) % MOD;
  return mom[n];
}
void solve() {
  long long i, j, k, x, n;
  cin >> N;
  for (i = 1; i <= N; i++) mom[i] = i;
  for (i = 1; i <= N; i++) {
    cin >> k;
    while (k--) {
      cin >> j >> x;
      n = find(j);
      mom[n] = i;
      V[n] = (V[j] + x) % MOD;
      ans = (ans + V[n]) % MOD;
    }
  }
  cout << (ans < 0 ? ans + MOD : ans) << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}
inline long long __gcd(long long a, long long b) {
  if (a == 0 || b == 0) {
    return a + b;
  }
  long long tempa, tempb;
  while (1) {
    if (a % b == 0)
      return b;
    else {
      tempa = a;
      tempb = b;
      a = tempb;
      b = tempa % tempb;
    }
  }
}
void pdash(long long n) {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < 30; j++) {
      cout << "-";
    }
    cout << "\n";
  }
}
long long power(long long x, long long y) {
  long long result = 1;
  while (y > 0) {
    if (y & 1) {
      result = (result * x);
    }
    y = y >> 1;
    x = (x * x);
  }
  return result;
}
long long power(long long x, long long y, long long z) {
  long long result = 1;
  x = x % z;
  while (y > 0) {
    if (y & 1) {
      result = (result * x) % z;
    }
    y = y >> 1;
    x = (x * x) % z;
  }
  return result;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
long long nCrF(long long n, long long r, long long p) {
  if (r == 0) return 1;
  vector<long long> f(n + 1);
  f[0] = 1;
  for (long long i = 1; i <= n; i++) f[i] = f[i - 1] * i % p;
  return (f[n] * modInverse(f[r], p) % p * modInverse(f[n - r], p) % p) % p;
}
vector<long long> cordinate_compression(vector<long long>& v) {
  vector<long long> p = v;
  make_uniquee(p);
  for (long long i = 0; i < (long long)((v).size()); i++)
    v[i] = (long long)(lower_bound(p.begin(), p.end(), v[i]) - p.begin());
  return p;
}
long long myrand(long long mod) { return rng() % mod; }
void make_uniquee(vector<long long>& vec) {
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
}
long long bitcount(long long u) {
  long long cnt = 0;
  while (u) {
    u = u & (u - 1);
    cnt++;
  }
  return cnt;
}
void svec(vector<long long>& vec) {
  cout << "{\n";
  for (long long i = 0; i < (long long)((vec).size()); i++)
    cout << "[" << i << "]  = " << vec[i] << "\n";
  cout << "}\n";
}
void sset(set<long long>& st) {
  cout << "{\n";
  for (auto& j : st) {
    cout << j << "\n";
  }
  cout << "}\n";
}
void smap(map<long long, long long>& mp) {
  cout << "{\n";
  for (auto& j : mp) cout << "[" << j.first << "]  = " << j.second << "\n";
  cout << "}\n";
}