#include <bits/stdc++.h>
using namespace std;
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3& p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3& p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3& p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3& p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4& p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4& p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4& p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2& p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2& p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2& p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2& p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long Gcd(long long v, long long b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return Gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return Gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long>& l,
                                     const pair<long long, long long>& r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long>& l,
                                     const pair<long long, long long>& r) {
  return {l.first - r.first, l.second - r.second};
}
ostream& operator<<(ostream& os, const vector<long long>& VEC) {
  for (auto v : VEC) os << v << " ";
  return os;
}
ostream& operator<<(ostream& os, const pair<long long, long long>& PAI) {
  os << PAI.first << " : " << PAI.second;
  return os;
}
ostream& operator<<(ostream& os, const pa3& PAI) {
  os << PAI.x << " : " << PAI.y << " : " << PAI.z;
  return os;
}
ostream& operator<<(ostream& os, const pa4& PAI) {
  os << PAI.x << " : " << PAI.y << " : " << PAI.z << " : " << PAI.w;
  return os;
}
ostream& operator<<(ostream& os,
                    const vector<pair<long long, long long> >& VEC) {
  for (auto v : VEC) os << v << " ";
  return os;
}
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long pr[1100000];
long long inv[1100000];
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = ((long long)pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
  }
}
long long dp[80][80][80];
long long dp2[80][80];
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 0; i < 80; i++)
    for (long long j = 0; j < 80; j++) dp2[i][j] = -1;
  dp2[0][0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < 80; j++)
      for (long long l = 0; l < 80; l++)
        for (long long e = 0; e < 80; e++) dp[l][j][e] = -1;
    dp[0][0][0] = 0;
    for (long long j = 0; j < m; j++) {
      long long y;
      cin >> y;
      for (long long l = 0; l <= m; l++)
        for (long long e = 0; e < k; e++)
          if (dp[j][l][e] >= 0) {
            dp[j + 1][l][e] = max(dp[j + 1][l][e], dp[j][l][e]);
            dp[j + 1][l + 1][(e + y) % k] =
                max(dp[j + 1][l + 1][(e + y) % k], dp[j][l][e] + y);
          }
    }
    vector<long long> a(k, -1);
    for (long long j = 0; j < k; j++)
      for (long long e = 0; e <= m / 2; e++) {
        a[j] = max(a[j], dp[m][e][j]);
      }
    for (long long j = 0; j < k; j++)
      if (dp2[i][j] >= 0) {
        for (long long l = 0; l < k; l++)
          if (a[l] >= 0) {
            dp2[i + 1][(j + l) % k] =
                max(dp2[i + 1][(j + l) % k], dp2[i][j] + a[l]);
          }
      }
  }
  cout << dp2[n][0] << endl;
}
signed main() {
  mod = 1000000007;
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n = 1;
  for (long long i = 0; i < n; i++) solve();
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given a matrix a of size n × m consisting of integers.

You can choose no more than \left⌊m/2\right⌋ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by k and this sum is the maximum.

In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by k.

Note that you can choose zero elements (and the sum of such set is 0).

Input

The first line of the input contains three integers n, m and k (1 ≤ n, m, k ≤ 70) — the number of rows in the matrix, the number of columns in the matrix and the value of k. The next n lines contain m elements each, where the j-th element of the i-th row is a_{i, j} (1 ≤ a_{i, j} ≤ 70).

Output

Print one integer — the maximum sum divisible by k you can obtain.

Examples

Input


3 4 3
1 2 3 4
5 2 2 2
7 1 1 4


Output


24


Input


5 5 4
1 2 4 2 1
3 5 1 2 4
1 5 7 1 2
3 8 7 1 2
8 4 7 1 6


Output


56

Note

In the first example, the optimal answer is 2 and 4 in the first row, 5 and 2 in the second row and 7 and 4 in the third row. The total sum is 2 + 4 + 5 + 2 + 7 + 4 = 24.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3& p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3& p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3& p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3& p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4& p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4& p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4& p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2& p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2& p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2& p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2& p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long Gcd(long long v, long long b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return Gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return Gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long& x, long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long>& l,
                                     const pair<long long, long long>& r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long>& l,
                                     const pair<long long, long long>& r) {
  return {l.first - r.first, l.second - r.second};
}
ostream& operator<<(ostream& os, const vector<long long>& VEC) {
  for (auto v : VEC) os << v << " ";
  return os;
}
ostream& operator<<(ostream& os, const pair<long long, long long>& PAI) {
  os << PAI.first << " : " << PAI.second;
  return os;
}
ostream& operator<<(ostream& os, const pa3& PAI) {
  os << PAI.x << " : " << PAI.y << " : " << PAI.z;
  return os;
}
ostream& operator<<(ostream& os, const pa4& PAI) {
  os << PAI.x << " : " << PAI.y << " : " << PAI.z << " : " << PAI.w;
  return os;
}
ostream& operator<<(ostream& os,
                    const vector<pair<long long, long long> >& VEC) {
  for (auto v : VEC) os << v << " ";
  return os;
}
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long pr[1100000];
long long inv[1100000];
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = ((long long)pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
  }
}
long long dp[80][80][80];
long long dp2[80][80];
void solve() {
  long long n, m, k;
  cin >> n >> m >> k;
  for (long long i = 0; i < 80; i++)
    for (long long j = 0; j < 80; j++) dp2[i][j] = -1;
  dp2[0][0] = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < 80; j++)
      for (long long l = 0; l < 80; l++)
        for (long long e = 0; e < 80; e++) dp[l][j][e] = -1;
    dp[0][0][0] = 0;
    for (long long j = 0; j < m; j++) {
      long long y;
      cin >> y;
      for (long long l = 0; l <= m; l++)
        for (long long e = 0; e < k; e++)
          if (dp[j][l][e] >= 0) {
            dp[j + 1][l][e] = max(dp[j + 1][l][e], dp[j][l][e]);
            dp[j + 1][l + 1][(e + y) % k] =
                max(dp[j + 1][l + 1][(e + y) % k], dp[j][l][e] + y);
          }
    }
    vector<long long> a(k, -1);
    for (long long j = 0; j < k; j++)
      for (long long e = 0; e <= m / 2; e++) {
        a[j] = max(a[j], dp[m][e][j]);
      }
    for (long long j = 0; j < k; j++)
      if (dp2[i][j] >= 0) {
        for (long long l = 0; l < k; l++)
          if (a[l] >= 0) {
            dp2[i + 1][(j + l) % k] =
                max(dp2[i + 1][(j + l) % k], dp2[i][j] + a[l]);
          }
      }
  }
  cout << dp2[n][0] << endl;
}
signed main() {
  mod = 1000000007;
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n = 1;
  for (long long i = 0; i < n; i++) solve();
  return 0;
}

```