#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
const int mod = 1000000007;
int n, k;
long long l;
int a[N];
int b[N];
int cnt[N];
int sum[N];
int len = 0;
int nxt[N];
long long good;
long long res = 0;
pair<int, int> c[N];
int id[N];
vector<vector<int> > d;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    c[i] = make_pair(a[i], i);
  }
  sort(c + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) id[c[i].second] = i;
  sort(a + 1, a + n + 1);
  nxt[n] = n;
  for (int i = n - 1; i >= 1; --i) {
    if (a[i] == a[i + 1])
      nxt[i] = nxt[i + 1];
    else
      nxt[i] = i;
  }
  k = min(k + 0ll, l / n + (l % n != 0 ? 1 : 0));
  d.resize(k + 5);
  for (int i = 0; i < k + 5; ++i) d[i].resize(n + 5);
  for (int i = 1; i <= n; ++i) d[1][i] = 1;
  for (int it = 2; it <= k; ++it) {
    sum[1] = d[it - 1][1];
    for (int i = 2; i <= n; ++i) sum[i] = (sum[i - 1] + d[it - 1][i]) % mod;
    for (int i = 1; i <= n; ++i) d[it][i] = sum[nxt[i]];
  }
  good = l / n;
  for (int cnt = 1; cnt <= k; ++cnt) {
    if (cnt <= good) {
      long long cur = 0;
      for (int i = 1; i <= n; ++i) cur = (cur + d[cnt][i]) % mod;
      cur = cur * 1ll * ((good - cnt + 1) % mod) % mod;
      res = (res + cur) % mod;
    }
    if (cnt - 1 <= good) {
      long long cur = 0;
      for (int i = 1; i <= l % n; ++i) cur = (cur + d[cnt][id[i]]) % mod;
      res = (res + cur) % mod;
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 = 1000005;
const int mod = 1000000007;
int n, k;
long long l;
int a[N];
int b[N];
int cnt[N];
int sum[N];
int len = 0;
int nxt[N];
long long good;
long long res = 0;
pair<int, int> c[N];
int id[N];
vector<vector<int> > d;
int main() {
  scanf("%d%I64d%d", &n, &l, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    c[i] = make_pair(a[i], i);
  }
  sort(c + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) id[c[i].second] = i;
  sort(a + 1, a + n + 1);
  nxt[n] = n;
  for (int i = n - 1; i >= 1; --i) {
    if (a[i] == a[i + 1])
      nxt[i] = nxt[i + 1];
    else
      nxt[i] = i;
  }
  k = min(k + 0ll, l / n + (l % n != 0 ? 1 : 0));
  d.resize(k + 5);
  for (int i = 0; i < k + 5; ++i) d[i].resize(n + 5);
  for (int i = 1; i <= n; ++i) d[1][i] = 1;
  for (int it = 2; it <= k; ++it) {
    sum[1] = d[it - 1][1];
    for (int i = 2; i <= n; ++i) sum[i] = (sum[i - 1] + d[it - 1][i]) % mod;
    for (int i = 1; i <= n; ++i) d[it][i] = sum[nxt[i]];
  }
  good = l / n;
  for (int cnt = 1; cnt <= k; ++cnt) {
    if (cnt <= good) {
      long long cur = 0;
      for (int i = 1; i <= n; ++i) cur = (cur + d[cnt][i]) % mod;
      cur = cur * 1ll * ((good - cnt + 1) % mod) % mod;
      res = (res + cur) % mod;
    }
    if (cnt - 1 <= good) {
      long long cur = 0;
      for (int i = 1; i <= l % n; ++i) cur = (cur + d[cnt][id[i]]) % mod;
      res = (res + cur) % mod;
    }
  }
  cout << res << endl;
  return 0;
}

```