Toad Zitz has an array of integers, each integer is between 0 and m-1 inclusive. The integers are a_1, a_2, …, a_n.

In one operation Zitz can choose an integer k and k indices i_1, i_2, …, i_k such that 1 ≤ i_1 < i_2 < … < i_k ≤ n. He should then change a_{i_j} to ((a_{i_j}+1) mod m) for each chosen integer i_j. The integer m is fixed for all operations and indices.

Here x mod y denotes the remainder of the division of x by y.

Zitz wants to make his array non-decreasing with the minimum number of such operations. Find this minimum number of operations.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) — the number of integers in the array and the parameter m.

The next line contains n space-separated integers a_1, a_2, …, a_n (0 ≤ a_i < m) — the given array.

Output

Output one integer: the minimum number of described operations Zitz needs to make his array non-decreasing. If no operations required, print 0.

It is easy to see that with enough operations Zitz can always make his array non-decreasing.

Examples

Input


5 3
0 0 0 1 2


Output


0


Input


5 7
0 6 1 3 2


Output


1

Note

In the first example, the array is already non-decreasing, so the answer is 0.

In the second example, you can choose k=2, i_1 = 2, i_2 = 5, the array becomes [0,0,1,3,3]. It is non-decreasing, so the answer is 1.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using pq = priority_queue<T, vector<T>, greater<T>>;
using ll = long long;
ll INF = LLONG_MAX / 4000 - 100;
ll mod = 1e9 + 7;
long long dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
vector<ll> prime;
class fact {
 public:
  long long fmod = 1e9 + 7;
  vector<long long> fac, finv, inv;
  fact(long long n, long long Mod = 1e9 + 7) {
    fmod = Mod;
    fac = vector<long long>(n + 1, 0);
    finv = vector<long long>(n + 1, 0);
    inv = vector<long long>(n + 1, 0);
    fac[0] = 1;
    for (long long i = 1; i < n + 1; i++) fac[i] = fac[i - 1] * i % fmod;
    for (long long i = 0; i < n + 1; i++) finv[i] = fact::POW(fac[i], fmod - 2);
    for (long long i = 0; i < n + 1; i++) inv[i] = POW(i, fmod - 2);
  }
  ll nCr(ll n, ll r) {
    if (n < r) return 0;
    return (fac[n] * finv[r] % fmod) * finv[n - r] % fmod;
  }
  ll POW(ll a, ll b) {
    ll c = 1;
    while (b > 0) {
      if (b & 1) {
        c = a * c % fmod;
      }
      a = a * a % fmod;
      b >>= 1;
    }
    return c;
  }
  inline long long operator[](long long i) { return fac[i]; }
};
void DEBUG(vector<long long> a) {
  for (long long i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << endl;
}
void EMP(long long x) { cout << "!!!" << x << "!!!" << endl; }
ll GCD(ll a, ll b) {
  ll c;
  while (b != 0) {
    c = a % b;
    a = b;
    b = c;
  }
  return a;
}
ll LCM(ll a, ll b) { return (a / GCD(a, b)) * (b / GCD(a, b)) * GCD(a, b); }
ll POW(ll a, ll b) {
  ll c = 1;
  while (b > 0) {
    if (b & 1) {
      c = a * c % mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return c;
}
void PRI(ll n) {
  bool a[n + 1LL];
  for (long long i = 0; i < n + 1LL; i++) {
    a[i] = 1LL;
  }
  for (long long i = 2; i < n + 1LL; i++) {
    if (a[i]) {
      prime.push_back(i);
      ll b = i;
      while (b <= n) {
        a[b] = 0;
        b += i;
      }
    }
  }
}
template <typename T>
T chmin(T& a, T b) {
  if (a > b) a = b;
  return a;
}
template <typename T>
T chmax(T& a, T b) {
  if (a < b) a = b;
  return b;
}
bool isSqrt(ll a) { return pow(sqrt(a), 2) == a ? 1 : 0; }
void YesNo(bool a) {
  if (a)
    cout << "Yes";
  else
    cout << "No";
  cout << endl;
}
void yesno(bool a) {
  if (a)
    cout << "yes";
  else
    cout << "no";
  cout << endl;
}
void YESNO(bool a) {
  if (a)
    cout << "YES";
  else
    cout << "NO";
  cout << endl;
}
double dis(long long x1, long long x2, long long y1, long long y2) {
  return sqrt((double)abs(x1 - x2) * (double)abs(x1 - x2) +
              (double)abs(y1 - y2) * (double)abs(y1 - y2));
}
long long ceili(long long x, long long y) {
  if (x % y == 0)
    return x / y;
  else
    return x / y + 1;
}
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;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (long long(i) = (0); (i) < (n); (i)++) cin >> a[i];
  long long l = -1, r = m - 1;
  while (r - l > 1) {
    long long mid = (r + l) / 2;
    long long last = 0;
    bool flag = true;
    for (long long(i) = (0); (i) < (n); (i)++) {
      if (a[i] <= last) {
        if (last - a[i] > mid) flag = false;
      } else {
        if (a[i] + mid < m) {
          last = a[i];
        } else {
          if ((a[i] + mid) % m < last) {
            last = a[i];
          }
        }
      }
    }
    if (flag)
      r = mid;
    else
      l = mid;
  }
  cout << r << endl;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  solve();
}
