489_F. Special Matrices
An n × n square matrix is special, if:

  * it is binary, that is, each cell contains either a 0, or a 1; 
  * the number of ones in each row and column equals 2. 



You are given n and the first m rows of the matrix. Print the number of special n × n matrices, such that the first m rows coincide with the given ones.

As the required value can be rather large, print the remainder after dividing the value by the given number mod.

Input

The first line of the input contains three integers n, m, mod (2 ≤ n ≤ 500, 0 ≤ m ≤ n, 2 ≤ mod ≤ 109). Then m lines follow, each of them contains n characters — the first rows of the required special matrices. Each of these lines contains exactly two characters '1', the rest characters are '0'. Each column of the given m × n table contains at most two numbers one.

Output

Print the remainder after dividing the required value by number mod.

Examples

Input

3 1 1000
011


Output

2


Input

4 4 100500
0110
1010
0101
1001


Output

1

Note

For the first test the required matrices are: 
    
    
      
    011  
    101  
    110  
      
    011  
    110  
    101  
    

In the second test the required matrix is already fully given, so the answer is 1.
```python
from math import *
from collections import *
import sys
sys.setrecursionlimit(10**9)
dp = [[-1 for i in range(505)] for j in range(505)]
def memo(row,z,o):
	if(row == n):
		if(z == 0 and o == 0):
			return 1
		else:
			return 0
	if(dp[z][o] != -1): return dp[z][o]
	
	if(z > 1):
		dp[z][o] += (z*(z-1)//2)*(memo(row+1,z-2,o+2))
		dp[z][o] %= mod
	if(z >= 1 and o >= 1):
		dp[z][o] += (z*o)*(memo(row+1,z-1,o))
		dp[z][o] %= mod
	if(o > 1):
		dp[z][o] += (o*(o-1)//2)*(memo(row+1,z,o-2))
		dp[z][o] %= mod
	#print(row,z,o,dp[z][o])
	dp[z][o] += 1
	dp[z][o] %= mod
	return dp[z][o]%mod

n,m,mod = map(int,input().split())
a = []
for i in range(m):
    s = list(input())
    a.append(s)
#print(a)
ct = [0 for i in range(n)]
for i in range(m):
	for j in range(n):
		if(a[i][j] == '1'):
			ct[j] += 1

z = ct.count(0)
o = ct.count(1)
ans = memo(m,z,o)
print(ans%mod)


```