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;
inline void Rd(int &res) {
  res = 0;
  char c;
  while (c = getchar(), c < 48)
    ;
  do res = (res << 1) + (res << 3) + (c ^ 48);
  while (c = getchar(), c >= 48);
}
inline void print(int x) {
  if (!x) return;
  print(x / 10);
  putchar((x % 10) ^ 48);
}
inline void sc(int x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  print(x);
  if (!x) putchar('0');
  putchar('\n');
}
inline void Max(int &x, int y) {
  if (x < y) x = y;
}
inline void Min(int &x, int y) {
  if (x > y) x = y;
}
const int M = 1e6 + 5;
const int P = 1e9 + 7;
int pos[M], sum[2][M], A[M], n, dp[2][M], en[M];
long long K, len;
struct node {
  int id, val;
} B[M];
bool cmp(node a, node b) {
  if (a.val != b.val) return a.val < b.val;
  return a.id > b.id;
}
int main() {
  scanf("%d", &n);
  cin >> len >> K;
  int x = min(K, len / n);
  int i, j, k;
  for (i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    B[i].val = A[i];
    B[i].id = i;
  }
  sort(B, B + n, cmp);
  sort(A, A + n);
  for (i = 0; i < n; i++) {
    pos[B[i].id] = i;
    en[i] = upper_bound(A, A + n, B[i].val) - A - 1;
  }
  int cur = 1, ans = 0;
  for (i = 1; i <= x; i++) {
    for (j = 0; j < n; j++) {
      if (i == 1)
        dp[cur][j] = 1;
      else
        dp[cur][j] = sum[cur ^ 1][en[j]];
      if (j - 1 >= 0)
        sum[cur][j] = (sum[cur][j - 1] + dp[cur][j]) % P;
      else
        sum[cur][j] = dp[cur][j];
      ans = (ans + 1ll * dp[cur][j] * ((len / n - i + 1) % P) % P) % P;
      if (B[j].id < len % n) ans = (ans + dp[cur][j]) % P;
    }
    cur ^= 1;
  }
  if (x < K) {
    for (j = 0; j < len % n; j++) {
      int y = pos[j];
      if (x)
        dp[cur][y] = sum[cur ^ 1][en[y]];
      else
        dp[cur][y] = 1;
      ans = (ans + dp[cur][y]) % P;
    }
  }
  printf("%d\n", ans);
  return 0;
}
