#include <bits/stdc++.h>
#pragma GCC optimize(2)
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 1000000007;
const int N = 1e6 + 10;
long long n, l, k;
long long a[N];
struct node {
  long long a, b;
  bool operator<(const node temp) const {
    if (a == temp.a)
      return b < temp.b;
    else
      return a < temp.a;
  }
};
long long sum[N];
long long dp[N];
int main() {
  n = read(), l = read(), k = read();
  vector<node> vec(n * k);
  for (int i = 0; i < n; i++) {
    a[i] = read();
    vec[i] = node{a[i], i};
  }
  for (int i = n; i < n * k; i++) {
    a[i] = a[i % n];
    vec[i] = node{a[i], i};
  }
  sort(vec.begin(), vec.end());
  long long ans = 0;
  for (int i = 0; i < n * k; i++) {
    long long temp = vec[i].a;
    long long pos = vec[i].b;
    if (pos < n)
      dp[pos] = 1;
    else
      dp[pos] = sum[pos / n - 1];
    sum[pos / n] += dp[pos];
    sum[pos / n] %= mod;
    long long dis = (l - pos - 1) / n + 1;
    dis %= mod;
    if (pos < l) ans = (ans + (dp[pos] * dis % mod)) % mod;
  }
  cout << ans << 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>
#pragma GCC optimize(2)
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int mod = 1000000007;
const int N = 1e6 + 10;
long long n, l, k;
long long a[N];
struct node {
  long long a, b;
  bool operator<(const node temp) const {
    if (a == temp.a)
      return b < temp.b;
    else
      return a < temp.a;
  }
};
long long sum[N];
long long dp[N];
int main() {
  n = read(), l = read(), k = read();
  vector<node> vec(n * k);
  for (int i = 0; i < n; i++) {
    a[i] = read();
    vec[i] = node{a[i], i};
  }
  for (int i = n; i < n * k; i++) {
    a[i] = a[i % n];
    vec[i] = node{a[i], i};
  }
  sort(vec.begin(), vec.end());
  long long ans = 0;
  for (int i = 0; i < n * k; i++) {
    long long temp = vec[i].a;
    long long pos = vec[i].b;
    if (pos < n)
      dp[pos] = 1;
    else
      dp[pos] = sum[pos / n - 1];
    sum[pos / n] += dp[pos];
    sum[pos / n] %= mod;
    long long dis = (l - pos - 1) / n + 1;
    dis %= mod;
    if (pos < l) ans = (ans + (dp[pos] * dis % mod)) % mod;
  }
  cout << ans << endl;
  return 0;
}

```