#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const int N = 1000010;
int n, m, nn, a[N], x[N], s[2][N];
long long l;
void GetData() {
  scanf("%d%I64d%ld", &n, &l, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    x[i] = a[i];
  }
}
void chkmod(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Add(int sum[], int p, int v) {
  for (; p <= nn; p += (p & -p)) chkmod(sum[p], v);
}
int Ask(int sum[], int p) {
  int res = 0;
  for (; p > 0; p -= (p & -p)) chkmod(res, sum[p]);
  return res;
}
int Calc(int x, int y) {
  long long t = l / n;
  if (t * n + y + 1 <= l) t++;
  t -= x - 1;
  return t % MOD;
}
void Solve() {
  int i, j, val, now, pre, ans = 0;
  sort(x, x + n);
  nn = unique(x, x + n) - x;
  for (i = 0; i < n; ++i) a[i] = lower_bound(x, x + nn, a[i]) - x + 1;
  for (now = 0, i = 1; i <= m; ++i) {
    pre = now;
    now ^= 1;
    for (j = 1; j <= nn; ++j) s[now][j] = 0;
    for (j = 0; j < n; ++j) {
      if (i == 1) {
        val = 1;
        Add(s[now], a[j], 1);
      } else {
        val = Ask(s[pre], a[j]);
        Add(s[now], a[j], val);
      }
      if ((i - 1) * n + j + 1 <= l)
        chkmod(ans, (long long)val * Calc(i, j) % MOD);
    }
  }
  printf("%d\n", ans);
}
int main() {
  GetData();
  Solve();
  return 0;
}

### Prompt

Your challenge is to write 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 MOD = 1000000007;
const int N = 1000010;
int n, m, nn, a[N], x[N], s[2][N];
long long l;
void GetData() {
  scanf("%d%I64d%ld", &n, &l, &m);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    x[i] = a[i];
  }
}
void chkmod(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void Add(int sum[], int p, int v) {
  for (; p <= nn; p += (p & -p)) chkmod(sum[p], v);
}
int Ask(int sum[], int p) {
  int res = 0;
  for (; p > 0; p -= (p & -p)) chkmod(res, sum[p]);
  return res;
}
int Calc(int x, int y) {
  long long t = l / n;
  if (t * n + y + 1 <= l) t++;
  t -= x - 1;
  return t % MOD;
}
void Solve() {
  int i, j, val, now, pre, ans = 0;
  sort(x, x + n);
  nn = unique(x, x + n) - x;
  for (i = 0; i < n; ++i) a[i] = lower_bound(x, x + nn, a[i]) - x + 1;
  for (now = 0, i = 1; i <= m; ++i) {
    pre = now;
    now ^= 1;
    for (j = 1; j <= nn; ++j) s[now][j] = 0;
    for (j = 0; j < n; ++j) {
      if (i == 1) {
        val = 1;
        Add(s[now], a[j], 1);
      } else {
        val = Ask(s[pre], a[j]);
        Add(s[now], a[j], val);
      }
      if ((i - 1) * n + j + 1 <= l)
        chkmod(ans, (long long)val * Calc(i, j) % MOD);
    }
  }
  printf("%d\n", ans);
}
int main() {
  GetData();
  Solve();
  return 0;
}

```