"""

奇数桁目の寄与を+,偶数を-として、mod 11で0なら、11の倍数

消して行けるのは…

A B C

 A + B + C = 0
-A + C = 0

より… A=C で無ければOK

A*2 + B = 0 にならないように、Bを置いていく問題になった

dp[付着？][和mod11]

"""

import sys
from sys import stdin

N = int(stdin.readline())

X = list(map(int,list(stdin.readline()[:-1])))

dp = [[[0] * 11 for i in range(2)] for j in range(2)]
dp[0][1][0] = 1

mod = 998244353


for i in range(N):

    ndp = [[[0] * 11 for i in range(2)] for j in range(2)]

    for flag in range(2):

        for d in range(10):

            if flag == 1 and X[i] < d:
                continue
            elif flag == 1 and X[i] == d:
                newflag = 1
            else:
                newflag = 0

            if i % 2 == 1:
                d *= -1

            for dig_puted in range(2):
                for oldd in range(11):

                    if (oldd*2 + d) % 11 == 0 and dig_puted != 0:
                        continue

                    newdigp = dig_puted
                    if d != 0:
                        newdigp |= 1

                    ndp[newdigp][newflag][(oldd+d)%11] += dp[dig_puted][flag][oldd]
                    ndp[newdigp][newflag][(oldd+d)%11] %= mod

    dp = ndp
    #print (dp)
    
ans = 0
for i in range(2):
    ans += dp[1][i][0]

print (ans % mod)