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, last, ans;
int a[1234567], p[1234567], dp[1234567][2];
long long l, rd;
bool cmp(int x, int y) {
  if (a[x] < a[y]) return true;
  return false;
}
void cal(int i, int j) {
  long long t = (rd - 1LL * j + 1LL) % 1000000007;
  if (i > last && last != 0) t = (t - 1LL + 1000000007) % 1000000007;
  long long tmp = (t * (1LL * dp[i][j & 1])) % 1000000007;
  ans = (ans + tmp) % 1000000007;
}
void init() {
  cin >> n >> l >> k;
  rd = l / (1LL * n);
  last = int(l % (1LL * n));
  if (last > 0) rd++;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    p[i] = i;
    dp[i][1] = 1;
    cal(i, 1);
  }
  sort(p + 1, p + n + 1, cmp);
}
void solve() {
  k = (int)min(1LL * k, rd);
  for (int j = 2; j <= k; j++) {
    int sum = 0, sta = j & 1, x = 1;
    for (int i = 1; i <= n; i++) {
      int tmp = a[p[i]];
      while (x <= n && a[p[x]] <= tmp)
        sum = (sum + dp[p[x++]][sta ^ 1]) % 1000000007;
      dp[p[i]][sta] = sum;
      cal(p[i], j);
    }
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
}
