#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 100;
const int INF = 1e9 + 19;
const int BASE = 1e9 + 7;
int n, k;
long long l;
int a[N];
pair<int, int> b[N];
int ans;
int csum[N], cnt[N], nom[N];
int add(int x, int y) {
  x += y;
  if (x >= BASE) x -= BASE;
  return x;
}
void go(int k) {
  for (int i = 0; i < n; i++) {
    long long len;
    if (i < l % n) {
      len = l / n + 1;
    } else {
      len = l / n;
    }
    len = len - k + 1;
    if (len < 0) len = 0;
    int cur = ((len % BASE) * 1LL * cnt[i]) % BASE;
    ans = add(ans, cur);
  }
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) b[i] = make_pair(a[i], i);
  sort(b, b + n);
  for (int i = 0; i < n; i++)
    nom[i] = upper_bound(b, b + n, make_pair(a[i], n)) - b;
  for (int i = 0; i < n; i++) cnt[i] = 1;
  ans = 0;
  for (int j = 1; j <= k; j++) {
    go(j);
    if (j == k) break;
    csum[0] = 0;
    for (int i = 0; i < n; i++) {
      csum[i + 1] = add(csum[i], cnt[b[i].second]);
    }
    for (int i = 0; i < n; i++) {
      cnt[i] = csum[nom[i]];
    }
  }
  cout << ans << endl;
}

### Prompt

Please formulate a CPP solution to 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 = 2e6 + 100;
const int INF = 1e9 + 19;
const int BASE = 1e9 + 7;
int n, k;
long long l;
int a[N];
pair<int, int> b[N];
int ans;
int csum[N], cnt[N], nom[N];
int add(int x, int y) {
  x += y;
  if (x >= BASE) x -= BASE;
  return x;
}
void go(int k) {
  for (int i = 0; i < n; i++) {
    long long len;
    if (i < l % n) {
      len = l / n + 1;
    } else {
      len = l / n;
    }
    len = len - k + 1;
    if (len < 0) len = 0;
    int cur = ((len % BASE) * 1LL * cnt[i]) % BASE;
    ans = add(ans, cur);
  }
}
int main() {
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) b[i] = make_pair(a[i], i);
  sort(b, b + n);
  for (int i = 0; i < n; i++)
    nom[i] = upper_bound(b, b + n, make_pair(a[i], n)) - b;
  for (int i = 0; i < n; i++) cnt[i] = 1;
  ans = 0;
  for (int j = 1; j <= k; j++) {
    go(j);
    if (j == k) break;
    csum[0] = 0;
    for (int i = 0; i < n; i++) {
      csum[i + 1] = add(csum[i], cnt[b[i].second]);
    }
    for (int i = 0; i < n; i++) {
      cnt[i] = csum[nom[i]];
    }
  }
  cout << ans << endl;
}

```