#include <bits/stdc++.h>
long long int n, m;
long long int p;
long long int ans;
struct st {
  int x, loc;
  bool operator<(const st& r) const { return x < r.x; }
};
st a[1001000];
long long int dp[1001000];
long long int dpb[1001000];
long long int mod = 1000000007;
int main() {
  int i, j, k;
  long long int t;
  scanf("%I64d%I64d%I64d", &n, &p, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i].loc = i;
  }
  std::sort(a, a + n);
  dpb[0] = 1;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      dp[j] = dpb[j];
      dpb[j] = 0;
    }
    k = 0;
    dpb[0] = 0;
    for (j = 0; j < n; j++) {
      if (j != 0) dpb[j] = dpb[j - 1];
      while (k < n && a[k].x == a[j].x) {
        dpb[j] += dp[k];
        dpb[j] %= mod;
        k++;
      }
      t = p / n - i;
      if (a[j].loc < p % n) {
        t++;
      }
      if (t < 0) t = 0;
      t %= mod;
      ans += (dpb[j] * t) % mod;
      ans %= mod;
    }
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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>
long long int n, m;
long long int p;
long long int ans;
struct st {
  int x, loc;
  bool operator<(const st& r) const { return x < r.x; }
};
st a[1001000];
long long int dp[1001000];
long long int dpb[1001000];
long long int mod = 1000000007;
int main() {
  int i, j, k;
  long long int t;
  scanf("%I64d%I64d%I64d", &n, &p, &m);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    a[i].loc = i;
  }
  std::sort(a, a + n);
  dpb[0] = 1;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      dp[j] = dpb[j];
      dpb[j] = 0;
    }
    k = 0;
    dpb[0] = 0;
    for (j = 0; j < n; j++) {
      if (j != 0) dpb[j] = dpb[j - 1];
      while (k < n && a[k].x == a[j].x) {
        dpb[j] += dp[k];
        dpb[j] %= mod;
        k++;
      }
      t = p / n - i;
      if (a[j].loc < p % n) {
        t++;
      }
      if (t < 0) t = 0;
      t %= mod;
      ans += (dpb[j] * t) % mod;
      ans %= mod;
    }
  }
  printf("%I64d", ans);
  return 0;
}

```