#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
int m, k, n, a[1000010], dp[1000010], ans = 0;
long long l;
pair<int, int> c[1000010];
int main() {
  m = read();
  l = read();
  k = read();
  n = m * k;
  for (int i = 0; i < m; ++i) a[i] = read();
  for (int i = 0; i < n; ++i) c[i].first = a[i % m], c[i].second = i;
  sort(c, c + n);
  long long tot = l / m;
  int yu = l % m;
  ans = l % 1000000007;
  for (int i = 0; i < n; ++i) {
    int id = c[i].second;
    if (id / m + 1 == 1) {
      dp[id / m + 1]++;
      continue;
    }
    dp[id / m + 1] += dp[id / m + 1 - 1], dp[id / m + 1] %= 1000000007;
    if (id >= l) continue;
    long long cnt = tot - (id / m + 1) + 1;
    if (id % m <= yu - 1) cnt++;
    ans += cnt % 1000000007 * dp[id / m + 1 - 1] % 1000000007;
    ans %= 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}

### 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;
inline char gc() {
  static char buf[1 << 16], *S, *T;
  if (S == T) {
    T = (S = buf) + fread(buf, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  long long x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
int m, k, n, a[1000010], dp[1000010], ans = 0;
long long l;
pair<int, int> c[1000010];
int main() {
  m = read();
  l = read();
  k = read();
  n = m * k;
  for (int i = 0; i < m; ++i) a[i] = read();
  for (int i = 0; i < n; ++i) c[i].first = a[i % m], c[i].second = i;
  sort(c, c + n);
  long long tot = l / m;
  int yu = l % m;
  ans = l % 1000000007;
  for (int i = 0; i < n; ++i) {
    int id = c[i].second;
    if (id / m + 1 == 1) {
      dp[id / m + 1]++;
      continue;
    }
    dp[id / m + 1] += dp[id / m + 1 - 1], dp[id / m + 1] %= 1000000007;
    if (id >= l) continue;
    long long cnt = tot - (id / m + 1) + 1;
    if (id % m <= yu - 1) cnt++;
    ans += cnt % 1000000007 * dp[id / m + 1 - 1] % 1000000007;
    ans %= 1000000007;
  }
  printf("%d\n", ans);
  return 0;
}

```