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 mxn = 1e6 + 100, mod = 1e9 + 7;
int n, k, last;
long long l, c, ans;
int dp[mxn][2], p[mxn], a[mxn];
inline void calc(int pos, int len) {
  long long block = (c - len + 1LL) % mod;
  if (pos > last && l % (long long)n) {
    block = (block - 1LL + mod) % mod;
  }
  long long xx = (1LL * block * (long long)dp[pos][len & 1]) % mod;
  ans = (ans + xx) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> l >> k;
  c = l / (long long)n;
  if (l % n) c++;
  k = (int)min((long long)k, c);
  last = (int)((l - 1LL) % (long long)n);
  ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    p[i] = i;
    dp[i][1] = 1;
    calc(i, 1);
  }
  sort(p, p + n, [](const int &i, const int &j) {
    return pair<int, int>(a[i], i) < pair<int, int>(a[j], j);
  });
  int po = 0;
  for (int j = 2; j <= k; j++) {
    po = 0;
    int x = j & 1, y = !x;
    int sum = 0;
    for (int rr = 0; rr < n; rr++) {
      int i = p[rr];
      while (po < n && a[p[po]] <= a[i]) {
        sum = (sum + dp[p[po++]][y]) % mod;
      }
      dp[i][x] = sum;
      calc(i, j);
    }
  }
  ans %= mod;
  ans = (ans + mod) % mod;
  cout << ans << "\n";
  return 0;
}
