You are given an integer m.

Let M = 2m - 1.

You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.

A set of integers S is called "good" if the following hold. 

  1. If <image>, then <image>. 
  2. If <image>, then <image>
  3. <image>
  4. All elements of S are less than or equal to M. 



Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.

Count the number of good sets S, modulo 109 + 7.

Input

The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).

The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.

Output

Print a single integer, the number of good sets modulo 109 + 7. 

Examples

Input

5 3
11010
00101
11000


Output

4


Input

30 2
010101010101010010101010101010
110110110110110011011011011011


Output

860616440

Note

An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}.
#include <bits/stdc++.h>
using namespace std;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 1e3 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 1e9 + 7;
char ss[52][maxn];
int cnt[maxn], fa[maxn], n, m, s[maxn][maxn], b[maxn];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
void link(int x, int y) {
  x = find(x), y = find(y);
  if (x != y) fa[x] = y, cnt[y] += cnt[x];
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= m; i++) scanf("%s", ss[i] + 1);
  for (int i = 1; i <= n; i++) fa[i] = i, cnt[i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      int bo = 1;
      for (int k = 1; k <= m; k++)
        if (ss[k][i] != ss[k][j]) bo = 0;
      if (bo) link(i, j);
    }
  s[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      s[i][j] = (1ll * j * s[i - 1][j] % mod + s[i - 1][j - 1]) % mod;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++) b[i] = (b[i] + s[i][j]) % mod;
  int ans = 1;
  for (int i = 1; i <= n; i++)
    if (fa[i] == i) ans = 1ll * ans * b[cnt[i]] % mod;
  write(ans);
  return 0;
}
