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 maxn = 1000000 + 23;
const long long MOD = 1e9 + 7, INF = (1 << 30);
int n, p, k, a[maxn], s[maxn], u[maxn], size;
long long ans, K, L, l, ele[maxn], seg[maxn], sum, F[maxn];
bool cmp(int x, int y) { return a[x] < a[y]; }
int num(int x, int y) { return (x - 1) * n + y; }
int main() {
  scanf("%d%I64d%I64d", &n, &l, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), u[i] = i;
  sort(u + 1, u + n + 1, cmp);
  size = 1, s[u[1]] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[u[i]] != a[u[i - 1]]) size++;
    s[u[i]] = size;
  }
  a[0] = INF;
  for (int i = 1; i <= n; i++) a[i] = s[i];
  for (int i = 1; i <= n; i++) F[i] = 1;
  for (long long i = 2; i <= K; i++) {
    p = 0, sum = 0;
    for (int j = 1; j <= n; j++) {
      while (a[u[p + 1]] <= a[u[j]]) {
        sum += F[num(i - 1, u[++p])];
        if (sum >= MOD) sum -= MOD;
      }
      F[num(i, u[j])] = sum;
    }
  }
  memset(seg, 0, sizeof(seg));
  memset(ele, 0, sizeof(ele));
  ans = 0, L = l / n;
  for (long long i = 1; i <= K; i++)
    for (int j = 1; j <= n; j++) {
      seg[i] += F[num(i, j)];
      if (seg[i] >= MOD) seg[i] -= MOD;
    }
  for (int i = 1; i <= n; i++)
    for (long long j = 1; j <= min(L + 1, K); j++) {
      ele[i] += F[num(j, i)];
      if (ele[i] >= MOD) ele[i] -= MOD;
    }
  for (long long i = 1; i <= min(K, L); i++)
    ans = (ans + seg[i] * ((L - i + 1) % MOD)) % MOD;
  for (long long i = L * n + 1; i <= l; i++) ans = (ans + ele[i - L * n]) % MOD;
  printf("%I64d\n", ans);
  return 0;
}
