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>.
#include <bits/stdc++.h>
using namespace std;
int n, k, aib[1000005], dp[1000005], a[1000005], ss[1000005];
long long l;
struct el {
  int val, poz;
  bool operator<(const el A) const { return val < A.val; }
} v[1000005];
inline void Fix(int &x) {
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
}
inline long long qry(int p) {
  int i;
  int sol = 0;
  for (i = p; i > 0; i -= (i & (-i))) {
    sol += aib[i];
    Fix(sol);
  }
  return sol;
}
inline void upd(int p, int val) {
  int i;
  for (i = p; i <= n; i += (i & (-i))) {
    aib[i] += val;
    Fix(aib[i]);
  }
}
inline void Normalize() {
  int i;
  sort(v + 1, v + n + 1);
  a[v[1].poz] = 1;
  for (i = 2; i <= n; ++i) {
    a[v[i].poz] = a[v[i - 1].poz];
    if (v[i].val > v[i - 1].val) ++a[v[i].poz];
  }
}
int main() {
  int i, j, t;
  long long buc, buc1;
  int sol = 0;
  cin.sync_with_stdio(0);
  cin >> n >> l >> k;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    v[i].val = a[i];
    v[i].poz = i;
  }
  Normalize();
  buc1 = l / n;
  if (l % n)
    buc = buc1 + 1;
  else
    buc = buc1;
  for (i = 1; i <= k && i <= buc; ++i) {
    for (j = 1; j <= n; ++j) ss[j] = dp[j];
    for (j = 1; j <= n; ++j) {
      if (i > 1)
        dp[j] = qry(a[j]);
      else
        dp[j] = 1;
      if (i <= buc1) {
        sol += (1LL * ((buc1 - i + 1) % 1000000007) * dp[j]) % 1000000007;
        Fix(sol);
      }
      if (j <= l - buc1 * n) {
        sol += dp[j];
        Fix(sol);
      }
    }
    for (j = 1; j <= n; ++j) aib[j] = 0;
    for (j = 1; j <= n; ++j) {
      upd(a[j], dp[j]);
    }
  }
  cout << sol;
  return 0;
}
