#include <bits/stdc++.h>
using namespace std;
long long i, j, k, l, n, m, s, an, K, X, tot = 1, oo, sum;
const long long pp = 1000000000 + 7;
const long long N = 2100000;
long long a[N], t[N];
long long tre[2][N];
long long f[N];
map<long long, long long> M;
void add(long long now, long long xx, long long vv) {
  for (long long i = xx; i <= tot; i += i & -i)
    tre[now][i] += vv, tre[now][i] %= pp;
}
long long find(long long now, long long xx) {
  long long s = 0;
  for (long long i = xx; i; i -= i & -i) s += tre[now][i], s %= pp;
  return s;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]), t[i] = a[i];
  sort(&t[1], &t[n + 1]);
  long long s = 1;
  for (long long i = 2; i <= n; i++)
    if (t[i] != t[s]) t[++s] = t[i];
  tot = s;
  for (long long i = 1; i <= s; i++) M[t[i]] = i;
  for (long long i = 1; i <= n; i++) a[i] = M[a[i]];
  int now = 0;
  tre[now][1] = 1;
  for (long long j = 1; j <= tot; j++)
    tre[now][j] += tre[now][j - 1], tre[now][j] %= pp;
  for (long long i = 1; i <= min(l, n * k); i++) {
    long long now = (i - 1) / n + 1;
    now &= 1;
    long long x = (i - 1) % n + 1;
    f[i] = tre[now ^ 1][a[x]];
    tre[now][a[x]] += f[i], tre[now][a[x]] %= pp;
    if (i / n != (i - 1) / n) {
      for (long long j = 1; j <= tot; j++) tre[now ^ 1][j] = 0;
      for (long long j = 1; j <= tot; j++)
        tre[now][j] += tre[now][j - 1], tre[now][j] %= pp;
    }
    long long s = ((l - i) / n + 1) % pp;
    an += f[i] * (s) % pp, an %= pp;
  }
  printf("%I64d\n", an);
  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;
long long i, j, k, l, n, m, s, an, K, X, tot = 1, oo, sum;
const long long pp = 1000000000 + 7;
const long long N = 2100000;
long long a[N], t[N];
long long tre[2][N];
long long f[N];
map<long long, long long> M;
void add(long long now, long long xx, long long vv) {
  for (long long i = xx; i <= tot; i += i & -i)
    tre[now][i] += vv, tre[now][i] %= pp;
}
long long find(long long now, long long xx) {
  long long s = 0;
  for (long long i = xx; i; i -= i & -i) s += tre[now][i], s %= pp;
  return s;
}
int main() {
  scanf("%I64d%I64d%I64d", &n, &l, &k);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]), t[i] = a[i];
  sort(&t[1], &t[n + 1]);
  long long s = 1;
  for (long long i = 2; i <= n; i++)
    if (t[i] != t[s]) t[++s] = t[i];
  tot = s;
  for (long long i = 1; i <= s; i++) M[t[i]] = i;
  for (long long i = 1; i <= n; i++) a[i] = M[a[i]];
  int now = 0;
  tre[now][1] = 1;
  for (long long j = 1; j <= tot; j++)
    tre[now][j] += tre[now][j - 1], tre[now][j] %= pp;
  for (long long i = 1; i <= min(l, n * k); i++) {
    long long now = (i - 1) / n + 1;
    now &= 1;
    long long x = (i - 1) % n + 1;
    f[i] = tre[now ^ 1][a[x]];
    tre[now][a[x]] += f[i], tre[now][a[x]] %= pp;
    if (i / n != (i - 1) / n) {
      for (long long j = 1; j <= tot; j++) tre[now ^ 1][j] = 0;
      for (long long j = 1; j <= tot; j++)
        tre[now][j] += tre[now][j - 1], tre[now][j] %= pp;
    }
    long long s = ((l - i) / n + 1) % pp;
    an += f[i] * (s) % pp, an %= pp;
  }
  printf("%I64d\n", an);
  return 0;
}

```