#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 10, M = 20, mo = 1e9 + 7;
vector<vector<long long> > f, sum;
vector<long long> g;
long long a[maxn], c[maxn];
long long n, l, k, K, B, ans, i, j, z, x, y, N, r;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  g.push_back(0);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    c[i] = a[i];
    g.push_back(0);
  }
  if (l <= n) {
    printf("%lld\n", l);
    return 0;
  }
  sort(c, c + n);
  K = min(l / n, k);
  N = (l / n) % mo;
  r = l % n;
  f.push_back(g);
  sum.push_back(g);
  for (i = 1; i <= k; i++) {
    f.push_back(g);
    sum.push_back(g);
  }
  for (i = 0; i < n; i++) {
    f[1][i] = 1;
    sum[0][i] = 1;
  }
  sum[1][0] = 1;
  for (i = 1; i < n; i++) sum[1][i] = (f[1][i] + sum[1][i - 1]) % mo;
  for (i = 2; i <= K; i++) {
    for (j = 0; j < n; j++) {
      z = upper_bound(c, c + n, c[j]) - c - 1;
      f[i][j] = sum[i - 1][z];
    }
    sum[i][0] = f[i][0];
    for (j = 1; j < n; j++) sum[i][j] = (f[i][j] + sum[i][j - 1]) % mo;
  }
  for (i = 1; i <= K; i++) {
    ans = (ans + (N - i + 1) * sum[i][n - 1]) % mo;
    for (j = 0; j < r; j++) {
      z = upper_bound(c, c + n, a[j]) - c - 1;
      ans = (ans + sum[i - 1][z]) % mo;
    }
  }
  if (r && K + 1 <= k) {
    i = K + 1;
    for (j = 0; j < r; j++) {
      z = upper_bound(c, c + n, a[j]) - c - 1;
      ans = (ans + sum[i - 1][z]) % mo;
    }
  }
  ans = (ans + mo) % mo;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 maxn = 1e6 + 10, M = 20, mo = 1e9 + 7;
vector<vector<long long> > f, sum;
vector<long long> g;
long long a[maxn], c[maxn];
long long n, l, k, K, B, ans, i, j, z, x, y, N, r;
int main() {
  scanf("%lld%lld%lld", &n, &l, &k);
  g.push_back(0);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    c[i] = a[i];
    g.push_back(0);
  }
  if (l <= n) {
    printf("%lld\n", l);
    return 0;
  }
  sort(c, c + n);
  K = min(l / n, k);
  N = (l / n) % mo;
  r = l % n;
  f.push_back(g);
  sum.push_back(g);
  for (i = 1; i <= k; i++) {
    f.push_back(g);
    sum.push_back(g);
  }
  for (i = 0; i < n; i++) {
    f[1][i] = 1;
    sum[0][i] = 1;
  }
  sum[1][0] = 1;
  for (i = 1; i < n; i++) sum[1][i] = (f[1][i] + sum[1][i - 1]) % mo;
  for (i = 2; i <= K; i++) {
    for (j = 0; j < n; j++) {
      z = upper_bound(c, c + n, c[j]) - c - 1;
      f[i][j] = sum[i - 1][z];
    }
    sum[i][0] = f[i][0];
    for (j = 1; j < n; j++) sum[i][j] = (f[i][j] + sum[i][j - 1]) % mo;
  }
  for (i = 1; i <= K; i++) {
    ans = (ans + (N - i + 1) * sum[i][n - 1]) % mo;
    for (j = 0; j < r; j++) {
      z = upper_bound(c, c + n, a[j]) - c - 1;
      ans = (ans + sum[i - 1][z]) % mo;
    }
  }
  if (r && K + 1 <= k) {
    i = K + 1;
    for (j = 0; j < r; j++) {
      z = upper_bound(c, c + n, a[j]) - c - 1;
      ans = (ans + sum[i - 1][z]) % mo;
    }
  }
  ans = (ans + mo) % mo;
  printf("%lld\n", ans);
  return 0;
}

```