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;
const int inf = 1e9 + 7;
struct node {
  int x;
  int ord;
  int r;
  int vis;
} a[1000005];
int cmp(node m, node n) { return m.x < n.x; }
int cmp1(node m, node n) { return m.ord < n.ord; }
int main() {
  long long n, k;
  long long l;
  cin >> n >> l >> k;
  vector<vector<long long> > dp(n + 3, vector<long long>(k + 2));
  long long i, j;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].x);
    a[i].ord = i;
  }
  long long p = l % n;
  if (l % n) {
    for (i = 1; i <= p; i++) a[i].vis = 1;
  }
  sort(a + 1, a + 1 + n, cmp);
  for (i = 1; i <= n; i++) {
    a[i].r = i;
  }
  for (j = 1; j <= n; j++) dp[j][1] = 1;
  for (i = 2; i <= k; i++) {
    int tot = 1;
    for (j = 1; j <= n; j++) {
      dp[j][i] = dp[j - 1][i];
      while (tot <= n && a[tot].x <= a[j].x) {
        dp[j][i] += dp[tot][i - 1];
        dp[j][i] %= inf;
        tot++;
      }
    }
  }
  long long ans = 0;
  long long len = l / n;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= k; j++) {
      if (a[i].vis == 1) {
        if (j <= len + 1) {
          ans += ((len + 1) + 1 - j) % inf * dp[i][j];
          ans %= inf;
        }
      } else {
        if (j <= len) {
          ans += (len + 1 - j) % inf * dp[i][j];
          ans %= inf;
        }
      }
    }
  }
  if (ans < 0) ans += inf;
  printf("%I64d\n", ans % inf);
}
