p02362 Single Source Shortest Path (Negative Edges)
Constraints

* 1 ≤ |V| ≤ 1000
* 0 ≤ |E| ≤ 2000
* -10000 ≤ di ≤ 10000
* There are no parallel edges
* There are no self-loops

Input

An edge-weighted graph G (V, E) and the source r.


|V| |E| r
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1


|V| is the number of vertices and |E| is the number of edges in G. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively. r is the source of the graph.

si and ti represent source and target vertices of i-th edge (directed) and di represents the cost of the i-th edge.

Output

If the graph contains a negative cycle (a cycle whose sum of edge costs is a negative value) which is reachable from the source r, print


NEGATIVE CYCLE


in a line.

Otherwise, print


c0
c1
:
c|V|-1


The output consists of |V| lines. Print the cost of the shortest path from the source r to each vertex 0, 1, ... |V|-1 in order. If there is no path from the source to a vertex, print "INF".

Examples

Input

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


Output

0
2
-3
-1


Input

4 6 0
0 1 2
0 2 3
1 2 -5
1 3 1
2 3 2
3 1 0


Output

NEGATIVE CYCLE


Input

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


Output

INF
0
-5
-3
```python
import collections
import sys
sys.setrecursionlimit(100000)

# ---- belmann-ford -----
V_num, E_num, r = [int(ele) for ele in input().split()]
adjc_list = []
edges = []
for _ in range(V_num):
    adjc_list.append([])

for _ in range(E_num):
    src, dst, w = [int(ele) for ele in input().split()]
    adjc_list[src].append([w, dst])
    edges.append([src, dst, w])

INF = 99999999999999
d = [INF for _ in range(V_num)]
d[r] = 0

# ---- dfs ----
stack = collections.deque()
yet = [ele for ele in range(V_num)]


def main():
    global edges
    # 連結成分の調査
    stack.clear()
    stack.appendleft(r)
    dfs(stack)
    # yet に残っているやつは非連結成分なので、bellman_fordの探索する辺リストから除外する
    edges = list(filter(lambda e: e[0] not in yet and e[1] not in yet, edges))
    output_ans() if bellman_ford() != -1 else print("NEGATIVE CYCLE")


def dfs(stack):
    if not stack:
        return

    n = stack[0]
    yet.remove(n)

    for _, dst in adjc_list[n]:  # stackの表面
        if dst in yet:  # 未訪問なら
            stack.appendleft(dst)
            dfs(stack)
    else:
        stack.popleft()


def bellman_ford():
    for _ in range(V_num - 1):
        for src, dst, w in edges:
            if d[dst] > w + d[src]:
                d[dst] = w + d[src]

    for src, dst, w in edges:
        if d[dst] > w + d[src]:
            return -1

    return


def output_ans():
    for d_i in d:
        if d_i == INF:
            print("INF")
        else:
            print(d_i)


if __name__ == "__main__":
    main()


```