#include <bits/stdc++.h>
using namespace std;
template <class T>
T nextInt() {
  T x = 0, p = 1;
  char c = 0;
  do {
    c = getchar();
  } while (c <= 32);
  if (c == '-') {
    p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * p;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
const long long P = (long long)1e9 + 7;
const int maxN = (int)2e6;
int n, k;
pair<int, int> a[maxN];
long long l, kBlocks;
long long& get(int i, int j) {
  static const int maxCnt = (int)4e6;
  static long long f[maxCnt] = {0};
  return f[(i + 1) * (k + 2) + j];
}
long long& sget(int i, int j) {
  static const int maxCnt = (int)4e6;
  static long long s[maxCnt] = {0};
  return s[(i + 1) * (k + 2) + j];
}
void calcS(int len) {
  sget(-1, len) = get(-1, len);
  for (int i = 0; i < n; i++) {
    sget(i, len) = (sget(i - 1, len) + get(i, len)) % P;
  }
}
long long g(int len, int last) {
  if (l % n == 0) {
    return (kBlocks - len + 1) % P;
  }
  if (a[last].second < l % n)
    return (kBlocks - len + 1) % P;
  else
    return (kBlocks - 1 - len + 1) % P;
}
int main() {
  n = nextInt<int>();
  l = nextInt<long long>();
  k = nextInt<int>();
  kBlocks = (l + n - 1) / n;
  for (int i = 0; i < n; i++) {
    a[i].first = nextInt<int>();
    a[i].second = i;
  }
  sort(a, a + n);
  get(-1, 0) = 1;
  calcS(0);
  for (int len = 1; len <= k; len++) {
    for (int last = 0; last < n; last++) {
      int value = a[last].first;
      int l = -1, r = n;
      while (r - l > 1) {
        int mid = (l + r) >> 1;
        if (a[mid].first <= value)
          l = mid;
        else
          r = mid;
      }
      get(last, len) = sget(l, len - 1);
    }
    calcS(len);
  }
  long long result = 0;
  for (int len = 1; len <= min<long long>(k, kBlocks); len++) {
    for (int last = 0; last < n; last++) {
      result = (result + g(len, last) * get(last, len)) % P;
    }
  }
  printf("%d\n", (int)result);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

While Duff was resting in the beach, she accidentally found a strange array b0, b1, ..., bl - 1 consisting of l positive integers. This array was strange because it was extremely long, but there was another (maybe shorter) array, a0, ..., an - 1 that b can be build from a with formula: bi = ai mod n where a mod b denoted the remainder of dividing a by b.

<image>

Duff is so curious, she wants to know the number of subsequences of b like bi1, bi2, ..., bix (0 ≤ i1 < i2 < ... < ix < l), such that: 

  * 1 ≤ x ≤ k
  * For each 1 ≤ j ≤ x - 1, <image>
  * For each 1 ≤ j ≤ x - 1, bij ≤ bij + 1. i.e this subsequence is non-decreasing. 



Since this number can be very large, she want to know it modulo 109 + 7.

Duff is not a programmer, and Malek is unavailable at the moment. So she asked for your help. Please tell her this number.

Input

The first line of input contains three integers, n, l and k (1 ≤ n, k, n × k ≤ 106 and 1 ≤ l ≤ 1018).

The second line contains n space separated integers, a0, a1, ..., an - 1 (1 ≤ ai ≤ 109 for each 0 ≤ i ≤ n - 1). 

Output

Print the answer modulo 1 000 000 007 in one line.

Examples

Input

3 5 3
5 9 1


Output

10


Input

5 10 3
1 2 3 4 5


Output

25

Note

In the first sample case, <image>. So all such sequences are: <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image>, <image> and <image>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
T nextInt() {
  T x = 0, p = 1;
  char c = 0;
  do {
    c = getchar();
  } while (c <= 32);
  if (c == '-') {
    p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * p;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
const long long P = (long long)1e9 + 7;
const int maxN = (int)2e6;
int n, k;
pair<int, int> a[maxN];
long long l, kBlocks;
long long& get(int i, int j) {
  static const int maxCnt = (int)4e6;
  static long long f[maxCnt] = {0};
  return f[(i + 1) * (k + 2) + j];
}
long long& sget(int i, int j) {
  static const int maxCnt = (int)4e6;
  static long long s[maxCnt] = {0};
  return s[(i + 1) * (k + 2) + j];
}
void calcS(int len) {
  sget(-1, len) = get(-1, len);
  for (int i = 0; i < n; i++) {
    sget(i, len) = (sget(i - 1, len) + get(i, len)) % P;
  }
}
long long g(int len, int last) {
  if (l % n == 0) {
    return (kBlocks - len + 1) % P;
  }
  if (a[last].second < l % n)
    return (kBlocks - len + 1) % P;
  else
    return (kBlocks - 1 - len + 1) % P;
}
int main() {
  n = nextInt<int>();
  l = nextInt<long long>();
  k = nextInt<int>();
  kBlocks = (l + n - 1) / n;
  for (int i = 0; i < n; i++) {
    a[i].first = nextInt<int>();
    a[i].second = i;
  }
  sort(a, a + n);
  get(-1, 0) = 1;
  calcS(0);
  for (int len = 1; len <= k; len++) {
    for (int last = 0; last < n; last++) {
      int value = a[last].first;
      int l = -1, r = n;
      while (r - l > 1) {
        int mid = (l + r) >> 1;
        if (a[mid].first <= value)
          l = mid;
        else
          r = mid;
      }
      get(last, len) = sget(l, len - 1);
    }
    calcS(len);
  }
  long long result = 0;
  for (int len = 1; len <= min<long long>(k, kBlocks); len++) {
    for (int last = 0; last < n; last++) {
      result = (result + g(len, last) * get(last, len)) % P;
    }
  }
  printf("%d\n", (int)result);
}

```