#include <bits/stdc++.h>
using namespace std;
const int Mod = int(1e9) + 7;
const int N = 1010000;
int m, n, k, a[N], ans, c[N];
long long l, dp[N], ss[N];
pair<int, int> p[N];
inline long long in() {
  int x = 0;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) a[i] = in();
  m = n * k;
  for (int i = 0; i < m; ++i) p[i] = make_pair(a[i % n], i);
  sort(p, p + m);
  for (int j = 0; j < m; ++j) {
    int i = p[j].second;
    if (i < n)
      dp[i] = 1;
    else
      dp[i] = ss[i / n - 1];
    ss[i / n] = (ss[i / n] + dp[i]) % Mod;
    if (i < l) {
      long long x = (l - 1 - i + n) / n;
      ans = (ans + (x % Mod) * dp[i]) % Mod;
    }
  }
  if (ans < 0) ans += Mod;
  printf("%d\n", ans);
}

### Prompt

Create a solution in cpp for 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 Mod = int(1e9) + 7;
const int N = 1010000;
int m, n, k, a[N], ans, c[N];
long long l, dp[N], ss[N];
pair<int, int> p[N];
inline long long in() {
  int x = 0;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 0; i < n; ++i) a[i] = in();
  m = n * k;
  for (int i = 0; i < m; ++i) p[i] = make_pair(a[i % n], i);
  sort(p, p + m);
  for (int j = 0; j < m; ++j) {
    int i = p[j].second;
    if (i < n)
      dp[i] = 1;
    else
      dp[i] = ss[i / n - 1];
    ss[i / n] = (ss[i / n] + dp[i]) % Mod;
    if (i < l) {
      long long x = (l - 1 - i + n) / n;
      ans = (ans + (x % Mod) * dp[i]) % Mod;
    }
  }
  if (ans < 0) ans += Mod;
  printf("%d\n", ans);
}

```