import argparse
parser = argparse.ArgumentParser(description="Compare two documents line by line")
parser.add_argument('a', help="reference document")
parser.add_argument('b', help="target document")
args = parser.parse_args()

fa = open(args.a,'r')
la = fa.read().split('\n')
fb = open(args.b,'r')
lb = fb.read().split('\n')

# Solve the Longest Common Subsequence (LCS) subproblem with dynamic programming
dp = [[0] * (len(lb)+1) for _ in range(len(la)+1)]
# dp[i+1][j+1] stores the LCS length between la[i] and lb[j]
for i in range(len(la)):
    for j in range(len(lb)):
        if la[i]==lb[j]:
            dp[i+1][j+1] = dp[i][j] + 1
        else:
            dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j])

# Backtrack to find one scheme to reduce both a and b to their LCS
diffs = []
i,j = len(la)-1, len(lb)-1
while i>=0 or j>=0: # current position (i+1, j+1), try to move to (0, 0)
    oi, oj = i, j
    while j >= 0 and dp[i+1][j+1]==dp[i+1][j]: # can safely delete lb[j]
        j -= 1
    while i >= 0 and dp[i+1][j+1]==dp[i][j+1]: # can safely delete la[i]
        i -= 1
    if i==oi and j==oj:
        assert dp[i+1][j+1]==dp[i][j]+1 and la[i]==lb[j]
        i,j = i-1,j-1
    else:
        diffs.append((oi,i,oj,j))
diffs.reverse()

def describe(oi,i,oj,j):
    # by diff convension line numbering starts from 1
    oi,i,oj,j = oi+1,i+1,oj+1,j+1
    def intv(x,y): # simplify expression if interval is one line
        return str(x) if x==y else f"{x},{y}"
    if i==oi:
        print(f"{i}a" + intv(j+1,oj))
    elif j==oj:
        print(intv(i+1,oi) + f"d{j}")
    else:
        print(intv(i+1,oi) + "c" + intv(j+1,oj))

for oi,i,oj,j in diffs:
    # delete (i,oi], add (j,oj]
    describe(oi,i,oj,j)
    for p in range(i+1,oi+1):
        print("< " + la[p])
    if oi>i and oj>j:
        print("---")
    for p in range(j+1,oj+1):
        print("> " + lb[p])