Problem: Rebel spy Heidi has just obtained the plans for the Death Star from the Empire and, now on her way to safety, she is trying to break the encryption of the plans (of course they are encrypted – the Empire may be evil, but it is not stupid!). The encryption has several levels of security, and here is how the first one looks.

Heidi is presented with a screen that shows her a sequence of integers A and a positive integer p. She knows that the encryption code is a single number S, which is defined as follows:

Define the score of X to be the sum of the elements of X modulo p.

Heidi is given a sequence A that consists of N integers, and also given an integer p. She needs to split A into 2 parts such that: 

  * Each part contains at least 1 element of A, and each part consists of contiguous elements of A. 
  * The two parts do not overlap. 
  * The total sum S of the scores of those two parts is maximized. This is the encryption code. 



Output the sum S, which is the encryption code.

Input

The first line of the input contains two space-separated integer N and p (2 ≤ N ≤ 100 000, 2 ≤ p ≤ 10 000) – the number of elements in A, and the modulo for computing scores, respectively.

The second line contains N space-separated integers which are the elements of A. Each integer is from the interval [1, 1 000 000].

Output

Output the number S as described in the problem statement.

Examples

Input

4 10
3 4 7 2


Output

16


Input

10 12
16 3 24 13 9 8 7 5 12 12


Output

13

Note

In the first example, the score is maximized if the input sequence is split into two parts as (3, 4), (7, 2). It gives the total score of <image>.

In the second example, the score is maximized if the first part consists of the first three elements, and the second part consists of the rest. Then, the score is <image>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 5e5 + 5;
const int maxk = 100 + 5;
int n, k, p;
int a[maxn];
int b[maxn];
int dp[2][maxn];
int fen[2][maxk];
void upd(int fen[], int p, int val) {
  p++;
  for (; p < maxk; p += p & -p) {
    chkmin(fen[p], val);
  }
}
int query(int fen[], int p) {
  p++;
  int res = INF;
  for (; p > 0; p -= p & -p) {
    chkmin(res, fen[p]);
  }
  return res;
}
void upd2(int fen[], int p, int val) {
  p++;
  for (; p > 0; p -= p & -p) {
    chkmin(fen[p], val);
  }
}
int query2(int fen[], int p) {
  p++;
  int res = INF;
  for (; p < maxk; p += p & -p) {
    chkmin(res, fen[p]);
  }
  return res;
}
void chemthan() {
  cin >> n >> k >> p;
  int t = 0;
  for (int i = (1); i < (n + 1); ++i) {
    cin >> a[i];
    b[i] = t = (t + a[i]) % p;
  }
  int pv = 0, nx = 1;
  for (int i = (0); i < (n + 1); ++i) dp[pv][i] = INF;
  dp[pv][0] = 0;
  for (int l = (0); l < (k); ++l) {
    for (int i = (0); i < (2); ++i) fill_n(fen[i], maxk, INF);
    for (int i = (0); i < (n + 1); ++i) dp[nx][i] = INF;
    for (int i = (1); i < (n + 1); ++i) {
      int r = b[i - 1];
      upd(fen[0], r, dp[pv][i - 1] - r);
      upd2(fen[1], r, dp[pv][i - 1] - r);
      int rr = b[i];
      chkmin(dp[nx][i], query(fen[0], rr) + rr);
      chkmin(dp[nx][i], query2(fen[1], rr + 1) + rr + p);
    }
    swap(pv, nx);
  }
  cout << dp[pv][n] << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}