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;
long long n, k, l;
long long mod = 1000000007LL, lInd[1000042];
vector<vector<long long> > ans, suff;
vector<long long> gen;
void gengen() {
  for (int i = 0; i < n + 5; i++) {
    gen.push_back(0);
  }
  for (int i = 0; i < k + 5; i++) {
    ans.push_back(gen);
    suff.push_back(gen);
  }
}
struct blq {
  long long val, ind;
} a[1000042];
long long ansE = 0, ansF = 0;
void solve() {
  for (long long num = k - 1; num >= 0; num--) {
    for (long long i = 0; i < n; i++) {
      long long ind = lInd[i];
      ans[num][i] += suff[num + 1][ind];
      ans[num][i] %= mod;
    }
    for (long long i = n - 1; i >= 0; i--) {
      suff[num][i] = ans[num][i] + suff[num][i + 1];
    }
  }
}
void clearr() {
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < k; j++) {
      ans[j][i] = 1;
    }
  }
}
bool cmp(blq u, blq v) { return u.val < v.val; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> l >> k;
  gengen();
  for (long long i = 0; i < n; i++) {
    cin >> a[i].val;
    a[i].ind = i;
  }
  sort(a, a + n, cmp);
  for (long long i = 0; i < n; i++) {
    long long j = i;
    for (; j >= 0 && a[i].val == a[j].val; j--) {
    }
    j++;
    lInd[i] = j;
  }
  if (l < k * n) {
    clearr();
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < n; j++) {
        if (i * n + a[j].ind >= l) {
          ans[i][j] = 0;
        }
      }
    }
    solve();
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < n; j++) {
        ansE += ans[i][j];
        ansE %= mod;
      }
    }
    cout << ansE << "\n";
    return 0;
  }
  clearr();
  for (long long i = 0; i < n; i++) {
    if (a[i].ind >= l % n) {
      ans[k - 1][i] = 0;
    }
  }
  solve();
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < k; j++) {
      ansE += ans[j][i];
      ansE %= mod;
    }
  }
  clearr();
  solve();
  for (long long i = 0; i < n; i++) {
    ansF += ans[0][i];
    ansF %= mod;
  }
  cout << (ansE + ((l / n - k + 1) % mod) * ansF % mod) % mod << "\n";
  return 0;
}
