For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.

Constraints

* $1 \leq n \leq 10,000$
* $0 \leq c_i \leq 100,000$
* $|E| < 500,000$
* All vertices are reachable from vertex $0$

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format:

$u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$

Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$).

Output

For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs.

Example

Input

5
0 3 2 3 3 1 1 2
1 2 0 2 3 4
2 3 0 3 3 1 4 1
3 4 2 1 0 1 1 4 4 3
4 2 2 1 3 3


Output

0 0
1 2
2 2
3 1
4 3
//dijkstra算法
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

//INFTY要足够大
const int INFTY = 1 << 21;

const int N = 10000;

vector<pair<int,int> >Graph[N];

int n;

void dijkstra(int u0) {
	int *d = new int[n];
	char *color = new char[n];
	for (int i=0; i<n; i++) {
		d[i] = INFTY;
		color[i] = 'W';
	}

	d[u0] = 0;
	color[u0] = 'G';
	priority_queue<pair<int,int> > PQ;
	PQ.push(make_pair(0,u0));

	while (!PQ.empty()) {
		pair<int,int> f = PQ.top();
		PQ.pop();
		int u = f.second;

		color[u] = 'B';

		//if (d[u] < f.first * -1) continue;

		for (int i=0; i<Graph[u].size(); i++) {
			pair<int,int> p = Graph[u][i];
			int v = p.first;
			if (color[v] != 'B'&& d[v] > d[u] + p.second) {
				d[v] = d[u] +p.second;
				color[v] = 'G';
				PQ.push(make_pair(d[v] * -1,v));
			}
		}
	}
	for (int i=0; i<n; i++) {
		cout << i << " " << (d[i] == INFTY ? -1 : d[i]) << endl;
	}
}

int main() {

	cin >> n;

	for (int i=0; i<n; i++) {
		int u,k;
		cin >> u >> k;

		for (int j = 0; j<k; j++) {
			int v,c;
			cin >> v >> c;
			Graph[u].push_back(make_pair(v,c));
		}
	}
	dijkstra(0);
	return 0;
}
