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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
const int MOD = 1000000007;
using namespace std;
template <typename T>
inline void read(T& x) {
  char ch;
  while (!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')))
    ;
  x = ch - '0';
  while (((ch = getchar()) >= '0') && (ch <= '9')) x = x * 10 + ch - '0';
}
int num[1001000];
int st[4001000];
int tmp[1001000];
void Init(int site, int L, int R) {
  if (L == R) {
    st[site] = 1;
    return;
  }
  int mid = (L + R) >> 1;
  Init(site << 1, L, mid);
  Init(site << 1 | 1, mid + 1, R);
  st[site] = st[site << 1] + st[site << 1 | 1];
}
int Q(int site, int L, int R, int goal) {
  if (R == goal) return st[site];
  int mid = (L + R) >> 1;
  if (goal <= mid) return Q(site << 1, L, mid, goal);
  return (st[site << 1] + Q(site << 1 | 1, mid + 1, R, goal)) % MOD;
}
void U(int site, int L, int R, int goal, int info) {
  if (L == R) {
    (st[site] += info) %= MOD;
    return;
  }
  int mid = (L + R) >> 1;
  if (goal <= mid)
    U(site << 1, L, mid, goal, info);
  else
    U(site << 1 | 1, mid + 1, R, goal, info);
  st[site] = (st[site << 1] + st[site << 1 | 1]) % MOD;
}
int main() {
  int n, k, i, j;
  long long l;
  scanf("%d %I64d %d", &n, &l, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &num[i]), tmp[i] = num[i];
  sort(tmp + 1, tmp + n + 1);
  for (i = 2, j = 1; i <= n; ++i)
    if (tmp[i] != tmp[j]) tmp[++j] = tmp[i];
  int m = j;
  for (i = 1; i <= n; ++i)
    num[i] = lower_bound(tmp + 1, tmp + m + 1, num[i]) - tmp;
  memset(st, 0, sizeof(int) * (m * 4 + 10));
  for (i = 1; i <= n; ++i) U(1, 1, m, num[i], 1);
  long long anw = l % MOD;
  k = min((long long)k, l / n + (l % n ? 1 : 0));
  for (i = 2; i <= k; ++i) {
    for (j = 1; j <= n; ++j) tmp[j] = Q(1, 1, m, num[j]);
    memset(st, 0, sizeof(int) * (m * 4 + 10));
    for (j = 1; j <= n; ++j) {
      U(1, 1, m, num[j], tmp[j]);
      if (j == l % n) {
        (anw += st[1]) %= MOD;
      }
    }
    (anw += (l / n - i + 1) % MOD * st[1]) %= MOD;
  }
  printf("%I64d\n", anw);
  return 0;
}
