Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10 ** 7)
read = sys.stdin.buffer.read 
inp = sys.stdin.buffer.readline
def inpS(): return inp().rstrip().decode()
readlines = sys.stdin.buffer.readlines 
MOD = 10**9+7

S = inpS()
N = len(S)
dp = [0]*13

dp[0] = 1
mul = 1
for i in range(N-1, -1, -1):
  nextDP = [0]*13
  s = S[i]
  if s == "?":
    for dgt in range(10):
      for mo in range(13):
        ind = (dgt*mul+mo)%13
        nextDP[ind] += dp[mo]
        nextDP[ind] %= MOD        
  else:
    dgt = int(S[i])
    for mo in range(13):
      ind = (dgt*mul+mo)%13
      nextDP[ind] += dp[mo]
      nextDP[ind] %= MOD    
  mul *= 10
  mul %= 13
  dp = nextDP
print(dp[5])    
      

Fixed submission:
import numpy as np
MOD = 10 ** 9 + 7

S = [int(x) if x != "?" else -1 for x in input()]
N = len(S)

# 現在考えてる一つ前の桁数で、余りが i になる整数の個数
r = [0]*13

# まず一つ目の値が１の位とする
s = S[0]
if s != -1:
  r[s] = 1
else:
  for i in range(10):
    r[i] = 1

# 桁を進めながら個数をカウント
for i in range(1, N):
    # dp[j]
    # 現在考えている桁(i)を除く、余りが j となる整数の個数
    dp = [0]*26
    # 桁が一つ上がるので、一つ前の余りが 10 倍になる
    for j in range(13):
      dp[j*10%13] += r[j] % MOD
    # アクセスを高速にするため２倍の長さとる (dp[i] == dp[i + 13])
    dp[13:26] = dp[0:13]
    r = [0]*13
    s = S[i]
    if s == -1:
      w = sum(dp[4:14])
      for j in range(13):
        r[j] += w
        w += dp[j+1] -dp[j+4]
    else:
      for j in range(13):
        r[j] += dp[13-s+j]
print(r[5]%MOD)

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10 ** 7)
read = sys.stdin.buffer.read 
inp = sys.stdin.buffer.readline
def inpS(): return inp().rstrip().decode()
readlines = sys.stdin.buffer.readlines 
MOD = 10**9+7

S = inpS()
N = len(S)
dp = [0]*13

dp[0] = 1
mul = 1
for i in range(N-1, -1, -1):
  nextDP = [0]*13
  s = S[i]
  if s == "?":
    for dgt in range(10):
      for mo in range(13):
        ind = (dgt*mul+mo)%13
        nextDP[ind] += dp[mo]
        nextDP[ind] %= MOD        
  else:
    dgt = int(S[i])
    for mo in range(13):
      ind = (dgt*mul+mo)%13
      nextDP[ind] += dp[mo]
      nextDP[ind] %= MOD    
  mul *= 10
  mul %= 13
  dp = nextDP
print(dp[5])    
      

Fixed submission:
import numpy as np
MOD = 10 ** 9 + 7

S = [int(x) if x != "?" else -1 for x in input()]
N = len(S)

# 現在考えてる一つ前の桁数で、余りが i になる整数の個数
r = [0]*13

# まず一つ目の値が１の位とする
s = S[0]
if s != -1:
  r[s] = 1
else:
  for i in range(10):
    r[i] = 1

# 桁を進めながら個数をカウント
for i in range(1, N):
    # dp[j]
    # 現在考えている桁(i)を除く、余りが j となる整数の個数
    dp = [0]*26
    # 桁が一つ上がるので、一つ前の余りが 10 倍になる
    for j in range(13):
      dp[j*10%13] += r[j] % MOD
    # アクセスを高速にするため２倍の長さとる (dp[i] == dp[i + 13])
    dp[13:26] = dp[0:13]
    r = [0]*13
    s = S[i]
    if s == -1:
      w = sum(dp[4:14])
      for j in range(13):
        r[j] += w
        w += dp[j+1] -dp[j+4]
    else:
      for j in range(13):
        r[j] += dp[13-s+j]
print(r[5]%MOD)

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10 ** 7)
read = sys.stdin.buffer.read 
inp = sys.stdin.buffer.readline
def inpS(): return inp().rstrip().decode()
readlines = sys.stdin.buffer.readlines 
MOD = 10**9+7

S = inpS()
N = len(S)
dp = [[0]*13 for _ in range(N+1)]

dp[0][0] = 1
for dgt in range(N):
  for mo in range(13):
    if S[dgt] == "?":
      for nxt in range(10):
        ind = (mo*10 + nxt) % 13
        dp[dgt+1][ind] += dp[dgt][mo]
        #dp[dgt+1][ind] %= MOD
    else:
      ind = (mo*10 + int(S[dgt])) % 13
      dp[dgt+1][ind] += dp[dgt][mo]
      #dp[dgt+1][ind] %= MOD
print(dp[N][5]%MOD)        

Fixed submission:
import numpy as np
MOD = 10 ** 9 + 7

S = [int(x) if x != "?" else -1 for x in input()]
N = len(S)

# 現在考えてる一つ前の桁数で、余りが i になる整数の個数
r = [0]*13

# まず一つ目の値が１の位とする
s = S[0]
if s != -1:
  r[s] = 1
else:
  for i in range(10):
    r[i] = 1

# 桁を進めながら個数をカウント
for i in range(1, N):
    # dp[j]
    # 現在考えている桁(i)を除く、余りが j となる整数の個数
    dp = [0]*26
    # 桁が一つ上がるので、一つ前の余りが 10 倍になる
    for j in range(13):
      dp[j*10%13] += r[j] % MOD
    # アクセスを高速にするため２倍の長さとる (dp[i] == dp[i + 13])
    dp[13:26] = dp[0:13]
    r = [0]*13
    s = S[i]
    if s == -1:
      w = sum(dp[4:14])
      for j in range(13):
        r[j] += w
        w += dp[j+1] -dp[j+4]
    else:
      for j in range(13):
        r[j] += dp[13-s+j]
print(r[5]%MOD)

Buggy python submission (verdict: Time Limit Exceeded):
import sys
sys.setrecursionlimit(10 ** 7)
read = sys.stdin.buffer.read 
inp = sys.stdin.buffer.readline
def inpS(): return inp().rstrip().decode()
readlines = sys.stdin.buffer.readlines 
MOD = 10**9+7

S = inpS()
N = len(S)
dp = [[0]*13 for _ in range(N+1)]

dp[0][0] = 1
for dgt in range(N):
  for mo in range(13):
    if S[dgt] == "?":
      for nxt in range(10):
        ind = (mo*10 + nxt) % 13
        dp[dgt+1][ind] += dp[dgt][mo]
        dp[dgt+1][ind] %= MOD
    else:
      ind = (mo*10 + int(S[dgt])) % 13
      dp[dgt+1][ind] += dp[dgt][mo]
      dp[dgt+1][ind] %= MOD
print(dp[N][5])        

Fixed submission:
import numpy as np
MOD = 10 ** 9 + 7

S = [int(x) if x != "?" else -1 for x in input()]
N = len(S)

# 現在考えてる一つ前の桁数で、余りが i になる整数の個数
r = [0]*13

# まず一つ目の値が１の位とする
s = S[0]
if s != -1:
  r[s] = 1
else:
  for i in range(10):
    r[i] = 1

# 桁を進めながら個数をカウント
for i in range(1, N):
    # dp[j]
    # 現在考えている桁(i)を除く、余りが j となる整数の個数
    dp = [0]*26
    # 桁が一つ上がるので、一つ前の余りが 10 倍になる
    for j in range(13):
      dp[j*10%13] += r[j] % MOD
    # アクセスを高速にするため２倍の長さとる (dp[i] == dp[i + 13])
    dp[13:26] = dp[0:13]
    r = [0]*13
    s = S[i]
    if s == -1:
      w = sum(dp[4:14])
      for j in range(13):
        r[j] += w
        w += dp[j+1] -dp[j+4]
    else:
      for j in range(13):
        r[j] += dp[13-s+j]
print(r[5]%MOD)