diff --git "a/03\354\243\274\354\260\250/\352\271\200\353\217\231\354\243\274/2807.py" "b/03\354\243\274\354\260\250/\352\271\200\353\217\231\354\243\274/2807.py"
new file mode 100644
index 0000000..46f4974
--- /dev/null
+++ "b/03\354\243\274\354\260\250/\352\271\200\353\217\231\354\243\274/2807.py"
@@ -0,0 +1,55 @@
+import collections
+import sys
+
+
+ALPHABETS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
+ALPHABET_LEN = len(ALPHABETS)
+
+
+def countCases(adjMat: dict[str, dict[str, int]], source: str, sink: str, *routes: list[str]) -> int:
+    if len(routes) < 2:
+        raise ValueError()
+
+    isPossible = True
+    ans = 1
+
+    # 인접행렬의 가중치가 크로스워드에 배치할 수 있는 단어의 개수라고 할 때,
+    # 한 번 배치할 때 마다 하나 씩 소모처리 해둠.
+    for route in routes:
+        isPossible &= adjMat[source][route] > 0
+        ans *= adjMat[source][route]
+        adjMat[source][route] -= 1
+
+        isPossible &= adjMat[route][sink] > 0
+        ans *= adjMat[route][sink]
+        adjMat[route][sink] -= 1
+
+    # 소모시켰던 가중치를 복구함으로서 인접행렬은 원상복구.
+    for route in routes:
+        adjMat[source][route] += 1
+        adjMat[route][sink] += 1
+
+    # 두 route가 다르면 대칭으로 뒤집어도 가능하니, 경우의 수도 두 배로...
+    ans *= len(set(routes))
+
+    return ans if isPossible else 0
+
+
+def solve(N: int, words: list[str]) -> int:
+    # Adjacent matrix
+    adjMat = collections.defaultdict(lambda: collections.defaultdict(lambda: 0))
+    answer = 0
+    for w in words:
+        adjMat[w[0]][w[-1]] += 1
+    for i in range(ALPHABET_LEN): # i = source
+        for j in range(ALPHABET_LEN): # j = sink
+            for k in range(ALPHABET_LEN): # k = route1
+                for l in range(k, ALPHABET_LEN): # l = route2
+                    answer += countCases(adjMat, ALPHABETS[i], ALPHABETS[j], ALPHABETS[k], ALPHABETS[l])
+    return answer
+
+
+if __name__ == "__main__":
+    N = int(sys.stdin.readline())
+    words = [sys.stdin.readline().strip() for _ in range(N)]
+    sys.stdout.write(str(solve(N, words)))