#include <bits/stdc++.h>
using namespace std;
const long long mod = (int)1e9 + 7;
struct node {
  long long v, i;
  bool f;
  bool operator<(node a) const { return v < a.v; }
} a[1000005];
long long i, i0, n, l, k, m, dp[1000005], ans[1000005], r[1000005], res = 0;
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  m = l % n;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i].v);
    a[i].f = (i < m), a[i].i = i;
  }
  sort(a, a + n);
  dp[0] = 1;
  for (i = 1; i <= min(k, l / n + 1); i++) {
    ans[i] = r[i] = 0;
    long long tmp = 0, p = 0;
    for (i0 = 0; i0 < n; i0++) {
      while (p < n && a[p].v <= a[i0].v) tmp += dp[p++];
      tmp %= mod;
      dp[i0] = tmp, ans[i] += dp[i0];
    }
    ans[i] %= mod;
    for (i0 = 0; i0 < n; i0++)
      if (!a[i0].f) r[i] += dp[i0];
    r[i] %= mod;
  }
  for (i = 1; i <= min(k, l / n + 1); i++) {
    res += ans[i] * ((l / n + 2 - i) % mod) % mod;
    res -= r[i];
    res %= mod, res += mod, res %= mod;
  }
  printf("%lld\n", res);
  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 long long mod = (int)1e9 + 7;
struct node {
  long long v, i;
  bool f;
  bool operator<(node a) const { return v < a.v; }
} a[1000005];
long long i, i0, n, l, k, m, dp[1000005], ans[1000005], r[1000005], res = 0;
int main() {
  scanf("%lld %lld %lld", &n, &l, &k);
  m = l % n;
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i].v);
    a[i].f = (i < m), a[i].i = i;
  }
  sort(a, a + n);
  dp[0] = 1;
  for (i = 1; i <= min(k, l / n + 1); i++) {
    ans[i] = r[i] = 0;
    long long tmp = 0, p = 0;
    for (i0 = 0; i0 < n; i0++) {
      while (p < n && a[p].v <= a[i0].v) tmp += dp[p++];
      tmp %= mod;
      dp[i0] = tmp, ans[i] += dp[i0];
    }
    ans[i] %= mod;
    for (i0 = 0; i0 < n; i0++)
      if (!a[i0].f) r[i] += dp[i0];
    r[i] %= mod;
  }
  for (i = 1; i <= min(k, l / n + 1); i++) {
    res += ans[i] * ((l / n + 2 - i) % mod) % mod;
    res -= r[i];
    res %= mod, res += mod, res %= mod;
  }
  printf("%lld\n", res);
  return 0;
}

```