'''
一次将边的两个端点并入并查集,如果某一次
并入两个节点之前发现两个节点已经在一个集合
中,则进行当前并操作后一定会形成环,找
第一个会形成环的边即可
'''
import sys
class MergeSetExt:
def __init__(self, max_key_val = 0, trans_key_func=None):
if trans_key_func is not None and max_key_val > 0:
# 如果能够提供转换key的回调,并查集内部用线性表存储
self.trans_key_callback = trans_key_func
self.m = [-1 for _ in range(max_key_val+1)]
self.__root2cluster_size = [0 for _ in range(max_key_val+1)]
else:
# 如果不能提供转换key的回调,并查集内部用hash表存储
self.trans_key_callback = None
self.m = {}
self.__root2cluster_size = {}
self.__root_cnt = 0
def getRoot(self, node):
buf = []
root = self.trans_key_callback(node) if self.trans_key_callback else node
while self.m[root] != root:
buf.append(root)
root = self.m[root]
for key in buf:
self.m[key] = root
return root
def merge(self, a, b):
orig_a, orig_b = a, b
if self.trans_key_callback:
a, b = self.trans_key_callback(a), self.trans_key_callback(b)
for node in [a, b]:
if (self.trans_key_callback is None and node not in self.m) or (self.trans_key_callback is not None and self.m[node] == -1):
self.m[node] = node
self.__root2cluster_size[node] = 1
self.__root_cnt += 1
root1 = self.getRoot(orig_a)
root2 = self.getRoot(orig_b)
if root1 != root2:
self.m[root1] = root2
self.__root2cluster_size[root2] += self.__root2cluster_size[root1]
if self.trans_key_callback:
self.__root2cluster_size[root1] = 0
else:
self.__root2cluster_size.pop(root1)
self.__root_cnt -= 1
def isInSameSet(self, a, b):
if a == b:
return True
orig_a, orig_b = a, b
if self.trans_key_callback:
a, b = self.trans_key_callback(a), self.trans_key_callback(b)
for node in [a, b]:
if self.m[node] == -1:
return False
else:
for node in [orig_a, orig_b]:
if node not in self.m:
return False
return self.getRoot(orig_a) == self.getRoot(orig_b)
n, m = map(int, input().split())
merge_set = MergeSetExt(max_key_val=n*n, trans_key_func=lambda x: (x[0]-1)*n+x[1])
ans = None
for i in range(1, m+1):
s = sys.stdin.readline()
a, b, dir = s.split()
a1, b1 = int(a), int(b)
a2, b2 = (a1, b1+1) if dir == 'R' else (a1+1, b1)
if merge_set.isInSameSet( (a1, b1), (a2, b2) ):
ans = i
break
else:
merge_set.merge( (a1, b1), (a2, b2) )
print(ans if ans is not None else 'draw')