#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int n, k;
long long l;
int a[1000001], d[1000001], acnt[1000001] = {};
int b[1000001], bcnt[1000001] = {};
vector<int> v;
inline void modAdd(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < (int)(n); ++i) scanf("%d", a + i);
  for (int i = 0; i < (int)(n); ++i) v.push_back(a[i]);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < (int)(n); ++i) b[i] = v[i];
  int sz = (int)v.size();
  for (int i = 0; i < (int)(n); ++i) a[i] = lower_bound(b, b + sz, a[i]) - b;
  long long gr = l / n;
  int grm = gr % MOD + MOD;
  int last = l - gr * n;
  for (int i = 0; i < (int)(last); ++i) b[i] = a[i];
  for (int i = 0; i < (int)(n); ++i) ++acnt[a[i]];
  for (int i = 0; i < (int)(last); ++i) ++bcnt[b[i]];
  d[0] = 1;
  int ans = 0;
  for (int i = 0; i < (int)(k); ++i) {
    for (int j = 1; j < sz; ++j) {
      modAdd(d[j], d[j - 1]);
    }
    for (int j = 0; j < (int)(sz); ++j) {
      ans = (ans + (long long)d[j] * bcnt[j]) % MOD;
    }
    if (i >= gr) break;
    int tot = 0;
    for (int j = 0; j < (int)(sz); ++j) {
      d[j] = (long long)d[j] * acnt[j] % MOD;
      modAdd(tot, d[j]);
    }
    ans = (ans + (long long)tot * (grm - i)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

### 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 = 1e9 + 7;
int n, k;
long long l;
int a[1000001], d[1000001], acnt[1000001] = {};
int b[1000001], bcnt[1000001] = {};
vector<int> v;
inline void modAdd(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < (int)(n); ++i) scanf("%d", a + i);
  for (int i = 0; i < (int)(n); ++i) v.push_back(a[i]);
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < (int)(n); ++i) b[i] = v[i];
  int sz = (int)v.size();
  for (int i = 0; i < (int)(n); ++i) a[i] = lower_bound(b, b + sz, a[i]) - b;
  long long gr = l / n;
  int grm = gr % MOD + MOD;
  int last = l - gr * n;
  for (int i = 0; i < (int)(last); ++i) b[i] = a[i];
  for (int i = 0; i < (int)(n); ++i) ++acnt[a[i]];
  for (int i = 0; i < (int)(last); ++i) ++bcnt[b[i]];
  d[0] = 1;
  int ans = 0;
  for (int i = 0; i < (int)(k); ++i) {
    for (int j = 1; j < sz; ++j) {
      modAdd(d[j], d[j - 1]);
    }
    for (int j = 0; j < (int)(sz); ++j) {
      ans = (ans + (long long)d[j] * bcnt[j]) % MOD;
    }
    if (i >= gr) break;
    int tot = 0;
    for (int j = 0; j < (int)(sz); ++j) {
      d[j] = (long long)d[j] * acnt[j] % MOD;
      modAdd(tot, d[j]);
    }
    ans = (ans + (long long)tot * (grm - i)) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

```