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
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 1e4 + 7, INF = 5e5 + 7;
int d[MAXN];
struct node {
	int id, dist;
	node(int id, int dist) :id(id), dist(dist) {}
	bool operator < (const node &rhs) const {
		return dist > rhs.dist;
	}
};
vector<pair<int, int> > G[MAXN];
priority_queue<node> nodes;
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
    	d[i] = INF;
	}
    for (int i = 0; i < n; i++) {
    	int u, k;
    	scanf("%d%d", &u, &k);
    	for (int j = 0; j < k; j++) {
    		int v, c;
    		scanf("%d%d", &v, &c);
    		G[u].push_back(make_pair(v, c));
		}
	}
	nodes.push(node(0, 0));
	while (!nodes.empty()) {
		node tmp = nodes.top();
		nodes.pop();
		if (tmp.dist > d[tmp.id]) continue;
		int cnt = G[tmp.id].size();
		for (int i = 0; i < cnt; i++) {
			if (d[G[tmp.id][i].first] > d[tmp.id] + G[tmp.id][i].second) {
				d[G[tmp.id][i].first] = d[tmp.id] + G[tmp.id][i].second;
				nodes.push(node(G[tmp.id][i].first, d[G[tmp.id][i].first]));
			}
		}
	}
	for (int i = 0; i < n; i++) {
		printf("%d %d\n", i, d[i]);
	}
	return 0;
}
