#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, mod = 1000000007;
long long rd() {
  long long x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
void ad(int &x, int y) { x += y, x -= x >= mod ? mod : 0; }
int n, m, kk, ans, a[N], b[N], cn[N], f[N];
vector<int> ps[N];
long long l;
int main() {
  n = rd(), l = rd() - 1, kk = rd();
  for (int i = 1; i <= n; ++i) a[i] = b[i] = rd();
  sort(b + 1, b + n + 1), m = unique(b + 1, b + n + 1) - b - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
    ps[a[i]].push_back(i - 1), ++cn[a[i]];
  }
  for (int i = 1; i <= kk; ++i) {
    int pr = i == 1;
    for (int j = 1; j <= m; ++j) {
      ad(pr, f[j]);
      f[j] = pr;
      for (int k = 0; k < cn[j]; ++k)
        ad(ans,
           max(0ll, (l - ps[j][k] - 1ll * (i - 2) * n) / n) % mod * f[j] % mod);
      f[j] = 1ll * f[j] * cn[j] % mod;
    }
  }
  printf("%d\n", 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>
using namespace std;
const int N = 1e6 + 10, mod = 1000000007;
long long rd() {
  long long x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
void ad(int &x, int y) { x += y, x -= x >= mod ? mod : 0; }
int n, m, kk, ans, a[N], b[N], cn[N], f[N];
vector<int> ps[N];
long long l;
int main() {
  n = rd(), l = rd() - 1, kk = rd();
  for (int i = 1; i <= n; ++i) a[i] = b[i] = rd();
  sort(b + 1, b + n + 1), m = unique(b + 1, b + n + 1) - b - 1;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
    ps[a[i]].push_back(i - 1), ++cn[a[i]];
  }
  for (int i = 1; i <= kk; ++i) {
    int pr = i == 1;
    for (int j = 1; j <= m; ++j) {
      ad(pr, f[j]);
      f[j] = pr;
      for (int k = 0; k < cn[j]; ++k)
        ad(ans,
           max(0ll, (l - ps[j][k] - 1ll * (i - 2) * n) / n) % mod * f[j] % mod);
      f[j] = 1ll * f[j] * cn[j] % mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}

```