Problem: Let's think about verbal arithmetic.

The material of this puzzle is a summation of non-negative integers represented in a decimal format, for example, as follows.


905 +  125 = 1030


In verbal arithmetic, every digit appearing in the equation is masked with an alphabetic character. A problem which has the above equation as one of its answers, for example, is the following.


ACM + IBM = ICPC


Solving this puzzle means finding possible digit assignments to the alphabetic characters in the verbal equation.

The rules of this puzzle are the following.

* Each integer in the equation is expressed in terms of one or more digits '0'-'9', but all the digits are masked with some alphabetic characters 'A'-'Z'.
* The same alphabetic character appearing in multiple places of the equation masks the same digit. Moreover, all appearances of the same digit are masked with a single alphabetic character. That is, different alphabetic characters mean different digits.
* The most significant digit must not be '0', except when a zero is expressed as a single digit '0'. That is, expressing numbers as "00" or "0123" is not permitted.



There are 4 different digit assignments for the verbal equation above as shown in the following table.

Mask| A| B| C| I| M| P
---|---|---|---|---|---|---
Case 1| 9| 2| 0| 1| 5| 3
Case 2| 9| 3| 0| 1| 5| 4
Case 3| 9| 6| 0| 1| 5| 7
Case 4| 9| 7| 0| 1| 5| 8

Your job is to write a program which solves this puzzle.

Input

The input consists of a number of datasets. The end of the input is indicated by a line containing a zero.

The number of datasets is no more than 100. Each dataset is formatted as follows.

> N
STRING 1
STRING 2
...
STRING N


The first line of a dataset contains an integer N  which is the number of integers appearing in the equation. Each of the following N  lines contains a string composed of uppercase alphabetic characters 'A'-'Z' which mean masked digits.

Each dataset expresses the following equation.

> STRING 1 + STRING 2 + ... + STRING N -1 = STRING N

The integer N  is greater than 2 and less than 13. The length of STRING i is greater than 0 and less than 9. The number of different alphabetic characters appearing in each dataset is greater than 0 and less than 11.

Output

For each dataset, output a single line containing the number of different digit assignments that satisfy the equation.

The output must not contain any superfluous characters.

Sample Input


3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0


Output for the Sample Input


4
1
8
30
0
0
0
40320






Example

Input

3
ACM
IBM
ICPC
3
GAME
BEST
GAMER
4
A
B
C
AB
3
A
B
CD
3
ONE
TWO
THREE
3
TWO
THREE
FIVE
3
MOV
POP
DIV
9
A
B
C
D
E
F
G
H
IJ
0


Output

4
1
8
30
0
0
0
40320

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef complex<double> Point;

#define PI acos(-1.0)
#define EPS 1e-10
const ll INF = 1e16;
const ll MOD = 1e9 + 7;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define rep(i, N) for (ll i = 0; i < (N); i++)
#define ALL(s) (s).begin(), (s).end()
#define EQ(a, b) (abs((a) - (b)) < EPS)
#define EQV(a, b) (EQ((a).real(), (b).real()) && EQ((a).imag(), (b).imag()))
#define fi first
#define se second
#define N_SIZE (1LL << 20)
#define NIL -1

int main()
{
    int n;
    while (cin >> n, n)
    {
        vector<string> s(n);
        rep(i, n) cin >> s[i];
        int al[26] = {}, v[26] = {}, cnt = 0;
        bool Top[26] = {};
        rep(i, 26) al[i] = -1;
        rep(i, n)
        {
            if (s[i].size() != 1)
                Top[s[i][0] - 'A'] = true;
            int tmp = 1;
            for (int j = s[i].size() - 1; j >= 0; j--)
            {
                if (al[s[i][j] - 'A'] == -1)
                    al[s[i][j] - 'A'] = cnt++;
                if (i != n - 1)
                    v[s[i][j] - 'A'] += tmp;
                else
                    v[s[i][j] - 'A'] -= tmp;
                tmp *= 10;
            }
        }
        int ans = 0;
        vector<int> perm(10);
        rep(i, 10) perm[i] = i;
        do
        {
            vector<int> tmp;
            for (int i = cnt; i < 10; i++)
                tmp.push_back(perm[i]);
            reverse(ALL(tmp));
            for (int i = cnt; i < 10; i++)
                perm[i] = tmp[i - cnt];
            bool f = false;
            int sum = 0;
            rep(i, 26)
            {
                if (al[i] == -1)
                    continue;
                int a = perm[al[i]];
                if (Top[i] && a == 0)
                {
                    f = true;
                    break;
                }
                sum += a * v[i];
            }
            if (!f && sum == 0)
                ans++;

        } while (next_permutation(ALL(perm)));
        cout << ans << endl;
    }
}