'''
转换成有源点，有汇点，流量上下界约束的最小流模型
用Dinic算法求解即可
'''
INF = 0x7fffffffffffffff
from collections import deque

class FortdFulkerson:
    # 输入图中所有边列表[(from1, to1, weight1), (form2, to2, weight2), ......]
    # 边可以有重边和自环边
    # souece_node 和 end_node 分别为源点和汇点
    # 所有节点从1开始连续编号, max_node_num是最大节点数，max_edge_nums是最大边数
    def __init__(self, edges, source_node, end_node, max_node_num, max_edge_num):
        self.edges = edges[::]
        self.source_node = source_node
        self.end_node = end_node
        self.max_edge_num = max_edge_num
        self.max_node_num = max_node_num

    #返回整个图的最大流和每条边的流量以及容量，(max_flow, [(from1, to1, flow1, weight1), (from1, to1, flow1, weight1), ......])
    def Dinic(self, rev_edge_w = None):                 # rev_edge_w是反向边边权，默认都是0
        e = [-1] * (self.max_edge_num*2 + 1)            # e[idx]表示编号为idx残量图边的终点, idx//2 就是残量图边对应的原图边的编号
        f = [-1] * (self.max_edge_num*2 + 1)            # f[idx]表示编号为idx的残量图边的流量
        ne = [-1] * (self.max_edge_num*2 + 1)           # ne[idx]表示根编号为idx的边同一个起点的下一条边的编号
        h = [-1] * (self.max_node_num + 1)              # h[a]表示节点a为起点的所有边的链表头对应的边的编号
        dis = [-1] * (self.max_node_num + 1)            # dis[a]表示点a到源点的距离，用于记录分层图信息
        cur = [-1] * (self.max_node_num + 1)            # cur[a]表示节点a在dfs搜索中第一次开始搜索的边的下标，也称当前弧，用于优化dfs速度
        orig_flow = [0] * (self.max_edge_num + 1)       # 原图中有向边的流量

        idx = 0
        for i, (a, b, w) in enumerate(self.edges):
            e[idx], f[idx], ne[idx], h[a] = b, w, h[a], idx
            idx += 1
            e[idx], f[idx], ne[idx], h[b] = a, 0, h[b], idx
            if rev_edge_w is not None:
                f[idx] = rev_edge_w[i]

            idx += 1

        # bfs搜索有没有增广路
        def bfs() -> bool:
            for i in range(self.max_node_num + 1):
                dis[i] = -1

            que = deque()
            que.append(self.source_node)
            dis[self.source_node] = 0
            cur[self.source_node] = h[self.source_node]

            while len(que) > 0:
                cur_node = que.popleft()
                idx = h[cur_node]
                while idx != -1:
                    next_node = e[idx]
                    if dis[next_node] == -1 and f[idx] > 0:
                        dis[next_node] = dis[cur_node] + 1
                        cur[next_node] = h[next_node]
                        if next_node == self.end_node:
                            return True

                        que.append(next_node)

                    idx = ne[idx]

            return False


        # dfs查找增广路, 返回当前残量图上node节点能流入汇点的不超过limit的最大流量有多事少
        def dfs(node, limit) -> int:
            if node == self.end_node:
                return limit

            flow = 0
            idx = cur[node]     # 从节点的当前弧开始搜索下一个点
            while idx != -1 and flow < limit:
                # 当前弧优化，记录每一个节点最后走的一条边，只要limit还没有减成0，已经搜过的边的终点
                # 能够汇入汇点的流量就已经全部用完了，另外一条路径到同一个点时候没必要重复搜索已经不会
                # 再提供流量贡献的邻接点
                cur[node] = idx

                next_node = e[idx]
                if dis[next_node] == dis[node]+1 and f[idx] > 0:
                    t = dfs(next_node, min(f[idx], limit - flow))
                    if t == 0:
                        # 已经无法提供流量的废点闪删除掉，不再参与搜索
                        dis[next_node] = -1

                    # 更新残量图边的流量
                    f[idx], f[idx^1], flow = f[idx]-t, f[idx^1]+t, flow+t

                    # 更新原图边的流量
                    if self.edges[idx>>1][0] == node:
                        orig_flow[idx>>1] += t
                    else:
                        orig_flow[idx>>1] -= t

                idx = ne[idx]

            return flow

        max_flow = 0
        while bfs():
            # 只要还有增广路，就dfs把增广路都找到，把增广路上的流量加到可行流上
            max_flow += dfs(self.source_node, INF)

        return max_flow, [(self.edges[i][0], self.edges[i][1], orig_flow[i], self.edges[i][2]) for i in range(len(self.edges))]

from collections import Counter
n, m, S, T = map(int, input().split())
SS, TT = n+1, n+2
edges = []
L = []
sum1, sum2 = Counter(), Counter()
for _ in range(m):
    a, b, l, u = map(int, input().split())
    edges.append((a, b, u-l))
    L.append(l)
    sum1[b] += l
    sum2[a] += l

# T和S间建立一条虚拟边
edges.append((T, S, INF))
sum1[S] += 0
sum2[T] += 0

for node in range(1, n+1):
    v1 = 0 if node not in sum1 else sum1[node]
    v2 = 0 if node not in sum2 else sum2[node]
    if v1 > v2:
        edges.append((SS, node, v1-v2))
    else:
        edges.append((node, TT, v2-v1))

algo = FortdFulkerson(edges, SS, TT, max_node_num=n+2, max_edge_num=len(edges))
ans = algo.Dinic()

F1 = 0
# 验证满流
flag = True
for a, b, f, w in ans[1]:
    if a == SS and f != w:
        flag = False
        break

    if w == INF:
        F1 = f

if not flag:
    print('No Solution')

else:
    # 再构建第二轮的图，求S到T的最大流，最后叠加结果
    e = []
    rev_w = []
    for a, b, f, w in ans[1]:
        if a >= 1 and a <= n and b >= 1 and b <= n and w != INF and w-f > 0:
            e.append((a, b, w-f))
            rev_w.append(f)     # 反向边的边权值
    algo = FortdFulkerson(e, T, S, max_node_num=n, max_edge_num=len(e))
    ans = algo.Dinic(rev_w)
    F2 = ans[0]
    ans = F1 - F2

    print(ans)