#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6;
const long long M = N;
const long long K = 1e6;
const long long MK = 1e6;
long long a[N], b[M], c[M], c1[M];
long long dp[MK];
long long m, k;
long long idx(long long i, long long x) {
  assert(0 <= i && i < m);
  assert(1 <= x && x <= k);
  return i * k + x - 1;
}
int main() {
  long long n, l;
  cin >> n >> l >> k;
  for (long long i = 0; i <= n - 1; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  long long j = 0;
  c[0] = 1;
  for (long long i = 1; i <= n - 1; i++) {
    if (b[i] == b[j]) {
      c[j]++;
    } else {
      j++;
      b[j] = b[i];
      c[j] = 1;
    }
  }
  m = j + 1;
  sort(a, a + (l % n));
  j = 0;
  for (long long i = 0; i <= (l % n) - 1; i++) {
    if (a[i] == b[j]) {
      c1[j]++;
    } else {
      assert(a[i] > b[j]);
      while (a[i] > b[j]) {
        j++;
      }
      assert(a[i] == b[j]);
      c1[j]++;
    }
  }
  for (long long i = 0; i <= m - 1; i++) {
    dp[idx(i, 1)] = c[i];
  }
  for (long long x = 2; x <= k; x++) {
    long long sum = 0;
    for (long long i = 0; i <= m - 1; i++) {
      sum = (sum + dp[idx(i, x - 1)]) % 1000000007LL;
      dp[idx(i, x)] = (c[i] * sum) % 1000000007LL;
    }
  }
  long long ans = 0;
  for (long long x = 1; x <= k; x++) {
    if ((l / n) >= x - 1) {
      if (x == 1) {
        for (long long i = 0; i <= m - 1; i++) {
          ans = (ans + c1[i]) % 1000000007LL;
        }
      } else {
        long long sum = 0;
        for (long long i = 0; i <= m - 1; i++) {
          sum = (sum + dp[idx(i, x - 1)]) % 1000000007LL;
          ans = (ans + c1[i] * sum) % 1000000007LL;
        }
      }
      if ((l / n) >= x) {
        for (long long i = 0; i <= m - 1; i++) {
          ans = (ans + (((l / n) - x + 1) % 1000000007LL) * dp[idx(i, x)]) %
                1000000007LL;
        }
      }
    }
  }
  cout << ans << '\n';
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const long long N = 1e6;
const long long M = N;
const long long K = 1e6;
const long long MK = 1e6;
long long a[N], b[M], c[M], c1[M];
long long dp[MK];
long long m, k;
long long idx(long long i, long long x) {
  assert(0 <= i && i < m);
  assert(1 <= x && x <= k);
  return i * k + x - 1;
}
int main() {
  long long n, l;
  cin >> n >> l >> k;
  for (long long i = 0; i <= n - 1; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  long long j = 0;
  c[0] = 1;
  for (long long i = 1; i <= n - 1; i++) {
    if (b[i] == b[j]) {
      c[j]++;
    } else {
      j++;
      b[j] = b[i];
      c[j] = 1;
    }
  }
  m = j + 1;
  sort(a, a + (l % n));
  j = 0;
  for (long long i = 0; i <= (l % n) - 1; i++) {
    if (a[i] == b[j]) {
      c1[j]++;
    } else {
      assert(a[i] > b[j]);
      while (a[i] > b[j]) {
        j++;
      }
      assert(a[i] == b[j]);
      c1[j]++;
    }
  }
  for (long long i = 0; i <= m - 1; i++) {
    dp[idx(i, 1)] = c[i];
  }
  for (long long x = 2; x <= k; x++) {
    long long sum = 0;
    for (long long i = 0; i <= m - 1; i++) {
      sum = (sum + dp[idx(i, x - 1)]) % 1000000007LL;
      dp[idx(i, x)] = (c[i] * sum) % 1000000007LL;
    }
  }
  long long ans = 0;
  for (long long x = 1; x <= k; x++) {
    if ((l / n) >= x - 1) {
      if (x == 1) {
        for (long long i = 0; i <= m - 1; i++) {
          ans = (ans + c1[i]) % 1000000007LL;
        }
      } else {
        long long sum = 0;
        for (long long i = 0; i <= m - 1; i++) {
          sum = (sum + dp[idx(i, x - 1)]) % 1000000007LL;
          ans = (ans + c1[i] * sum) % 1000000007LL;
        }
      }
      if ((l / n) >= x) {
        for (long long i = 0; i <= m - 1; i++) {
          ans = (ans + (((l / n) - x + 1) % 1000000007LL) * dp[idx(i, x)]) %
                1000000007LL;
        }
      }
    }
  }
  cout << ans << '\n';
}

```