There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes. You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin. The transaction takes D_i minutes for each gold coin you give. You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

* 2 \leq N \leq 50
* N-1 \leq M \leq 100
* 0 \leq S \leq 10^9
* 1 \leq A_i \leq 50
* 1 \leq B_i,C_i,D_i \leq 10^9
* 1 \leq U_i < V_i \leq N
* There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).
* Each city t=2,...,N can be reached from City 1 with some number of railroads.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N


Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Examples

Input

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5


Output

2
14


Input

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4


Output

5
5
7


Input

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1


Output

1
9003
14606
16510
16576


Input

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7


Output

1
3
5


Input

2 1 0
1 2 1 1
1 1000000000
1 1


Output

1000000001
# ダイクストラ
def main():
    import sys
    sys.setrecursionlimit(10**9)
    input = sys.stdin.readline
    from heapq import heappush, heappop

    N, M, S = map(int, input().split())
    S = min(S, 2500)
    graph = [[] for _ in range(N)]

    for u, v, a, b in (map(int, input().split()) for _ in range(M)):
        graph[u-1].append((v-1, a, b))
        graph[v-1].append((u-1, a, b))

    for i, (c, d) in enumerate(map(int, input().split()) for _ in range(N)):
        graph[i].append((i, -c, d))

    hq = [(0, 0, S)]
    INF = 10**18
    # dp[v][silver]
    # 都市vにsilverを使って到達したときの時間の最小値
    dp = [[INF]*2501 for _ in range(N)]
    dp[0][S] = 0

    while hq:
        time, v, silver = heappop(hq)
        if dp[v][silver] < time:
            continue

        for dest, a, b in graph[v]:
            next_s = silver-a if silver-a < 2500 else 2500
            if next_s < 0 or dp[dest][next_s] <= time+b:
                continue
            for i in range(next_s, -1, -1):
                if dp[dest][i] > time+b:
                    dp[dest][i] = time+b
                else:
                    break
            heappush(hq, (time+b, dest, next_s))

    print(*(min(dp[i]) for i in range(1, N)), sep='\n')

main()

