You have a rectangular n × m-cell board. Some cells are already painted some of k colors. You need to paint each uncolored cell one of the k colors so that any path from the upper left square to the lower right one doesn't contain any two cells of the same color. The path can go only along side-adjacent cells and can only go down or right.

Print the number of possible paintings modulo 1000000007 (109 + 7).

Input

The first line contains three integers n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10). The next n lines contain m integers each — the board. The first of them contains m uppermost cells of the board from the left to the right and the second one contains m cells from the second uppermost row and so on. If a number in a line equals 0, then the corresponding cell isn't painted. Otherwise, this number represents the initial color of the board cell — an integer from 1 to k.

Consider all colors numbered from 1 to k in some manner.

Output

Print the number of possible paintings modulo 1000000007 (109 + 7).

Examples

Input

2 2 4
0 0
0 0


Output

48


Input

2 2 4
1 2
2 1


Output

0


Input

5 6 10
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0


Output

3628800


Input

2 6 10
1 2 3 4 5 6
0 0 0 0 0 0


Output

4096
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(0.0) * 2.0;
const double eps = 1e-12;
const int step[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <class T>
inline T max1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a > b ? a : b;
  if (c != -1234567) ans = max1(ans, c);
  if (d != -1234567) ans = max1(ans, d);
  return ans;
}
template <class T>
inline T min1(T a, T b, T c = -1234567, T d = -1234567) {
  T ans = a < b ? a : b;
  if (c != -1234567) ans = min(ans, c);
  if (d != -1234567) ans = min(ans, d);
  return ans;
}
template <class T>
inline T gcd1(T a, T b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd1(b, a % b);
}
template <class T>
inline T lb(T num) {
  return num & (-num);
}
inline int jud(double a, double b) {
  if (abs1(a - b) < eps) return 0;
  if (a < b) return -1;
  return 1;
}
template <typename it, typename t1>
inline int find(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 1) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1 && be != 0) be--;
    if (!f_small && jud(*(a + be), val) == -1 && be != na - 1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1 && be != 0) be--;
    if (f_small && jud(*(a + be), val) == 1 && be != na - 1) be++;
  }
  return be;
}
inline int bitnum(unsigned long long nValue) {
  nValue = ((0xaaaaaaaaaaaaaaaaull & nValue) >> 1) +
           (0x5555555555555555ull & nValue);
  nValue = ((0xccccccccccccccccull & nValue) >> 2) +
           (0x3333333333333333ull & nValue);
  nValue = ((0xf0f0f0f0f0f0f0f0ull & nValue) >> 4) +
           (0x0f0f0f0f0f0f0f0full & nValue);
  nValue = ((0xff00ff00ff00ff00ull & nValue) >> 8) +
           (0x00ff00ff00ff00ffull & nValue);
  nValue = ((0xffff0000ffff0000ull & nValue) >> 16) +
           (0x0000ffff0000ffffull & nValue);
  nValue = ((0xffffffff00000000ull & nValue) >> 32) +
           (0x00000000ffffffffull & nValue);
  return nValue;
}
long long pow(long long n, long long m, long long mod = 0) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
const int mod = 1000000007;
const int maxn = 11;
int n, m, nc, rannc, nxtnc;
int have[maxn][maxn], cant[maxn][maxn], mp[maxn][maxn];
int used[maxn];
long long dfs(int i, int j) {
  if (j == m) j = 0, i++;
  if (i == n) return 1;
  have[i][j] = 0;
  if (i != 0) have[i][j] |= have[i - 1][j];
  if (j != 0) have[i][j] |= have[i][j - 1];
  if ((have[i][j] == (1 << nc) - 1)) return 0;
  if (mp[i][j]) {
    if ((1 << mp[i][j] - 1) & have[i][j]) return 0;
    have[i][j] |= 1 << (mp[i][j] - 1);
    return dfs(i, j + 1);
  }
  long long rans = 0;
  int nxtnc1 = nxtnc;
  for (int i1 = 0; i1 < nc; i1++)
    if (((cant[i][j] & (1 << i1)) == 0) && (have[i][j] & (1 << i1)) == 0) {
      if (!used[i1] && i1 != nxtnc) continue;
      have[i][j] |= (1 << i1);
      used[i1]++;
      if (used[i1] == 1) {
        for (; nxtnc < nc; nxtnc++)
          if (used[nxtnc] == 0) break;
        rannc--;
        rans += ((rannc + 1) * dfs(i, j + 1)) % mod;
        rans %= mod;
        rannc++;
        nxtnc = nxtnc1;
      } else {
        rans += dfs(i, j + 1);
        rans %= mod;
      }
      used[i1]--;
      have[i][j] ^= (1 << i1);
    }
  return rans;
}
int main() {
  scanf("%d%d%d", &n, &m, &nc);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) scanf("%d", mp[i] + j);
  if (n + m - 1 > nc) {
    cout << 0 << endl;
    return 0;
  }
  rannc = nc;
  for (int i = n - 1; i >= 0; i--)
    for (int j = m - 1; j >= 0; j--) {
      if (i != n - 1) {
        if (mp[i + 1][j]) cant[i][j] |= 1 << mp[i + 1][j] - 1;
        cant[i][j] |= cant[i + 1][j];
      }
      if (j != m - 1) {
        if (mp[i][j + 1]) cant[i][j] |= 1 << mp[i][j + 1] - 1;
        cant[i][j] |= cant[i][j + 1];
      }
      if (mp[i][j]) {
        if ((1 << (mp[i][j] - 1)) & cant[i][j]) {
          cout << 0 << endl;
          return 0;
        }
        used[mp[i][j] - 1]++;
        if (used[mp[i][j] - 1] == 1) rannc--;
      }
    }
  for (; used[nxtnc]; nxtnc++)
    ;
  cout << dfs(0, 0) << endl;
  return 0;
}
