#include <bits/stdc++.h>
using namespace std;
int n, k;
long long l;
int a[1010101];
vector<int> v;
unordered_map<int, int> z;
long long d[1010101];
long long s[1010101];
long long w;
long long pot(long long n, int k) {
  if (k == 0) return 1;
  if (k % 2 == 0) {
    long long u = pot(n, k / 2);
    return (u * u) % 1000000007;
  } else {
    return (pot(n, k - 1) * n) % 1000000007;
  }
}
long long inv(long long x) { return pot(x, 1000000007 - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (i == n - 1 || v[i] != v[i + 1]) {
      c++;
      z[v[i]] = c;
    }
  }
  long long x = l % n;
  long long e1 = l / n;
  long long e2 = l / n + 1;
  e1 %= 1000000007;
  e2 %= 1000000007;
  int cc = 0;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= c; j++) {
      d[j] = 0;
    }
    for (int j = 0; j < n; j++) {
      cc++;
      if (cc > l) goto lol;
      long long u = s[z[a[j]]];
      if (i == 1) u = 1;
      d[z[a[j]]] += u;
      d[z[a[j]]] %= 1000000007;
      if (j < x)
        w += u * e2;
      else
        w += u * e1;
      w %= 1000000007;
    }
    for (int j = 1; j <= c; j++) {
      s[j] = s[j - 1] + d[j];
      s[j] %= 1000000007;
    }
    e1--;
    if (e1 < 0) e1 += 1000000007;
    e2--;
    if (e2 < 0) e2 += 1000000007;
  }
lol:
  cout << w << "\n";
}

### 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;
int n, k;
long long l;
int a[1010101];
vector<int> v;
unordered_map<int, int> z;
long long d[1010101];
long long s[1010101];
long long w;
long long pot(long long n, int k) {
  if (k == 0) return 1;
  if (k % 2 == 0) {
    long long u = pot(n, k / 2);
    return (u * u) % 1000000007;
  } else {
    return (pot(n, k - 1) * n) % 1000000007;
  }
}
long long inv(long long x) { return pot(x, 1000000007 - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> l >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (i == n - 1 || v[i] != v[i + 1]) {
      c++;
      z[v[i]] = c;
    }
  }
  long long x = l % n;
  long long e1 = l / n;
  long long e2 = l / n + 1;
  e1 %= 1000000007;
  e2 %= 1000000007;
  int cc = 0;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= c; j++) {
      d[j] = 0;
    }
    for (int j = 0; j < n; j++) {
      cc++;
      if (cc > l) goto lol;
      long long u = s[z[a[j]]];
      if (i == 1) u = 1;
      d[z[a[j]]] += u;
      d[z[a[j]]] %= 1000000007;
      if (j < x)
        w += u * e2;
      else
        w += u * e1;
      w %= 1000000007;
    }
    for (int j = 1; j <= c; j++) {
      s[j] = s[j - 1] + d[j];
      s[j] %= 1000000007;
    }
    e1--;
    if (e1 < 0) e1 += 1000000007;
    e2--;
    if (e2 < 0) e2 += 1000000007;
  }
lol:
  cout << w << "\n";
}

```