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
from heapq import heappop,heappush
INF = float("inf")
white = 0; gray = 1; black =2

def dijkstra(s):
    # 頂点sをrootに設定
    dist[s] = 0 
    heappush(PQ,(0,s))
    color[s] = gray

    while PQ:
        #最短経路木に頂点(最短経路木とつながる最小コストの辺を持つ頂点)を追加
        mincost,u = heappop(PQ)
        color[u] = black
        if dist[u] < mincost: continue # 最短じゃなかったら無視

        # 追加した頂点uの隣接頂点についてsからの最小コストを更新
        for v,cost in adj[u]:
            if color[v] == black: continue # 確定済み
            if dist[v] > dist[u] + cost:
                dist[v] = dist[u] + cost
                heappush(PQ,(dist[v],v))
                color[v] = gray

if __name__ == "__main__":
    n = int(input())
    adj = [[] for _ in range(n)]
    for i in range(n):
        inp = [int(x) for x in input().split()]
        u = inp[0]
        for v,c in zip(inp[2::2],inp[3::2]): adj[u].append((v,c))
    color = [white]*n #各頂点の訪問状態 （white:未訪問 gray:暫定 black:確定）
    dist = [INF]*n # 始点sからの最小コスト
    PQ = []
    dijkstra(0)
    for i in range(n): print("{} {}".format(i,dist[i]))
