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 = 1e6 + 5;
const int mod = 1000000007;
long long dp[3][maxn];
pair<int, int> pp[maxn];
map<int, int> mp1, mp2;
long long fastpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = res * x % mod;
    }
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  int n, k;
  long long l;
  scanf("%d%I64d%d", &n, &l, &k);
  long long p = l / n, q = l % n;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &pp[i].first);
    pp[i].second = i;
    ++mp1[pp[i].first];
    if (i < q) {
      ++mp2[pp[i].first];
    }
  }
  long long ans = 0;
  sort(pp, pp + n);
  long long *now = dp[0], *nxt = dp[1];
  dp[0][0] = dp[1][0] = 1;
  for (int i = 0; i < n; ++i) {
    if (i && pp[i].first == pp[i - 1].first) continue;
    for (int j = 1; j <= min(p + 1, (long long)k); ++j) {
      nxt[j] = nxt[j - 1] * mp1[pp[i].first] % mod;
      if (p >= j) {
        ans = (ans + (p - j + 1) % mod * nxt[j] % mod) % mod;
      }
      if (p >= j - 1 && pp[i].second < q) {
        ans = (ans + nxt[j - 1] * mp2[pp[i].first] % mod) % mod;
      }
      nxt[j] = (nxt[j] + now[j]) % mod;
    }
    swap(now, nxt);
  }
  printf("%I64d\n", ans);
  return 0;
}
